
Citation 
 Permanent Link:
 http://digital.auraria.edu/AA00001803/00001
Material Information
 Title:
 Threedimensional classical solid geometry
 Creator:
 Hatoum, Rached T
 Publication Date:
 1992
 Language:
 English
 Physical Description:
 xi, 128 leaves : illustrations ; 29 cm
Subjects
 Subjects / Keywords:
 Geometry, Solid ( lcsh )
Geometry, Solid ( fast )
 Genre:
 bibliography ( marcgt )
theses ( marcgt ) nonfiction ( marcgt )
Notes
 Bibliography:
 Includes bibliographical references.
 General Note:
 Submitted in partial fulfillment of the requirements for the degree, Master of Science, [Department of] Applied Mathematics.
 Statement of Responsibility:
 by Rached T. Hatoum.
Record Information
 Source Institution:
 University of Colorado Denver
 Holding Location:
 Auraria Library
 Rights Management:
 All applicable rights reserved by the source institution and holding location.
 Resource Identifier:
 26891528 ( OCLC )
ocm26891528
 Classification:
 LD1190.L622 1992m .H37 ( lcc )

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THREEDIMENSIONAL CLASSICAL SOLID GEOMETRY by Rached T. Hatoum B.s., Metropolitan State College at Denver, 1990 A thesis submitted to the Faculty of the Graduate School of the University of Colorado at Denver in partial fulfillment of the requirements for the degree of Master of Science Applied Mathematics 1992
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This thesis for the Master of Science degree by Rached T. Hatoum has been approved for the Department of Mathematics by Stanley E. Payne William E. Cherowitzo Zenas R. Hartvigson Date
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Hatoum, Rached T. (M.S., Applied Mathematics) ThreeDimensional Classical Solid Geometry Thesis directed by Professor Stanley E. Payne ABSTRACT This thesis work consists of six chapters that can be considered as a detailed study of the threedimensional classical solid geometry. It contains short and to the point specific definitions, theorems and their detailed proofs, problems with thorough solutions for each chapter and general problems that deal with all the material presented in this thesis. During my research for this work, I noticed that the definitions and theorems were shuffled throughout each book I read. So, I decided to organize my thesis into chapters where I listed each definition and each theorem in their proper place in each chapter. Each definition and each theorem were used later in their proper place in the same chapter or a different chapter. Sometimes, I wrote some lemmas that I used in proving and solving some theorems and general problems that I formulated. As a whole, I feel that I have accomplished a reorganization with everything in its place. I also that this work satisfied my curiosity in seeing the concepts of the classical threedimensional solid geometry in a different iii
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perspective from the analytical ways used in dealing with solid geometry problems in calculus. The concepts in this thesis are encountered by most of the students nowadays throughout their college education while studying calculus with analytic geometry. so all the details they encounter analytically are dealt with using both threedimensional and twodimensional geometries. As I mentioned before, the whole work is very organized, and it is very beneficial for undergraduate and graduate students because of the great connection with calculus and analytic geometry used in our modern education. This abstract accurately represents the content of the candidate's thesis. I recommend its publication. Signed Stanley E. Payne iv
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Acknowledgement This entire thesis work is a body of knowledge that has been done before by Euclid, and then translated into many languages. Euclid's work in geometry has been rewritten by many mathematicians over the ages, and to those mathematicians I owe the information collected in this v
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Contents Chapter 1. Determination Of A Plane 2. 1.1 Definition Of A Plane 1.2 Representation Of A Plane 1.3 Properties Of The Plane 1.4 Axioms 1.5 Determination Of A Plane 1.5.1 Theorem 1.5.2 Theorem 1.5.3 Theorem 1.5.4 Definition 1.5.5 Theorem . 1.6 Conclusion .. .. .. .. 1.7 1.8 1.9 Relative Position Of Two Straight Lines .. .. .. .. Theorem . . .. Definition Of A Tetrahedron 1.9.1 Problem . 1. 9. 2 Problem Straight Line Parallel To A Plane 2.1 Necessary And Sufficient Condition For A Straight Line To Be Parallel To A Plane 2.1.1 Definition .. . 2.1.2 Theorem . .. . . vi 1 1 1 1 2 2 2 2 5 6 6 7 7 9 10 11 12 13 13 13 13
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i 3. 2.2 Axiom w .. .. .. .. 2.3 Theorem 2.4 Theorem 2.5 Theorem 2.6 Theorem 2.7 Theorem .. . 2.8 2.9 Theorem .. .. .. .. Angle Between Two Skew Straight Lines Parallel Planes 3.1 Necessary And Sufficient Condition For Two Planes To Be Parallel 3.1.1 Definition .. .. .. .. .. 3.1.2 Theorem .. .. .. .. .. 3.1.3 Necessary Condition 3.1.4 Sufficient Condition 3.2 Theorem .. .. .. .. .. .. 3.3 Theorem . . 3.4 Theorem . . . 3.5 Theorem . . . 3.6 Theorem . 3.7 Theorem . 3.8 Theorem . . . 3.9 Theorem . . . 3.10 Converse .. .. .. .. .. .. .. .. .. .. .. .. .. .. vii 15 15 16 17 18 20 20 22 23 23 23 23 23 24 25 27 28 29 30 32 33 34 36
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3.11 Some Problems On Chapters Two And Three 4. Perpendicular Straight Lines And Planes 4.1 Necessary And Sufficient Condition For A Straight Line To Be Perpendicular To A Plane 4.1.1 Definition 4 1. 2 Theorem 4.1.3 Necessary Condition 4.1.4 Sufficient Condition 4.2 Theorem Of Three Perpendiculars 4.3 Theorem 4.4 Theorem 4.5 Theorem 4.6 Corollary 4.7 Theorem 4.8 Corollary 4.9 Theorem 4.10 Theorem 4.11 Converse 4.12 Perpendicular And Obliques 4.12.1 Theorem 4.13 Converse 4.14 The Distance From A Point To A Plane 4.15 Symmetry With Respect To A Point viii 37 43 43 43 43 43 44 45 46 47 48 49 50 52 53 54 55 56 56 58 60 60
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4.16 Symmetry With Respect To A straight Line . 4.17 symmetry With Respect To A Plane 4.18 Theorem . 4.19 Axis Of A Circle 4.20 Theorem 4.21 Some Problems On Chapter Four 5. Perpendicular Planes 5.1 Definitions 5.1.1 Definition 5.1.2 Definition 5.1.3 Definition 5.1.4 Definition 5.2 The Plane Angle Of A Dihedral 5.3 Necessary And Sufficient Condition For Two Planes To Be Perpendicular 5.3.1 Definition 5.3.2 Theorem 5.3.3 Necessary Condition 5.3.4 Sufficient Condition 5.4 Theorem 5.5 Theorem 5.6 Theorem 5.7 Theorem 5.8 Theorem ix 61 61 61 62 62 63 68 68 68 68 68 69 69 70 70 70 70 71 72 73 74 75 76
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5.9 Theorem 5.10 Theorem 5.11 Some Problems On Chapter Five 6. Projections on A Plane 6.1 Projections Parallel To A Given Direction 6.1.1 Projection Of A Point 6.1.2 Projection Of A Figure 6.1.3 Theorem 6.1.4 Special Cases 6.1.5 Theorem 6.1.6 Theorem 6.2 Orthogonal Projection Of A Right Angle 6.2.1 Theorem 6.2.2 Necessary COndition 6.2.3 Sufficient Condition 6.2.4 Theorem 6.3 Angle Between A Straight Line And A Plane 77 77 78 80 80 80 80 81 82 82 83 84 84 84 86 87 88 6. 3 .1 Definition 88 6.3.2 Theorem 88 6.3.3 Length Of The Orthogonal Projection Of A Straight Line Segment On A Plane 89 6.3.4 Theorem 90 6.4 Orthogonal Projection Of A Polygonal Surface X . 91
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6.4.1 Theorem . .. . 91 6.4.2 Area Of The Orthogonal Projection Of Triangle ABC 91 6.4.3 Area Of The Orthogonal Projection Of Any Polygon 94 6.5 The Common Perpendicular 6.5.1 Definition 6.5.2 Theorem 6.5.3 Existence 6.5.4 AB Is The Only Common Perpendicular 6.5.5 AB Is The Shortest Segment 6.6 Some Problems on Chapter Six 7. General Problems Bibliography .. xi 95 95 95 95 96 97 98 104 128
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1. Determination Of A Plane 1.1 Definition Of A Plane A plane is a smooth surface such that if any two points on it are joined by a straight line, then this straight line will lie completely in the plane, every point on the straight line will lie in the plane. 1.2 Representation Of A Plane A plane is represented by a parallelogram or by an angle as in figures 1.1 and 1.2. A plane is designated by a letter such as P. N.B.: If A and Bare two points in plane P, then the straight line 1 joining A and B lies completely in P according to definition 1.1. I Figure 1.1 Figure 1.2 1.3 Properties Of The Plane a) A plane can be extended indefinitely, so the straight lines that form the parallelogram in figure 1.2 are not the borders of the plane P. The parallelogram is only a representation, and the same thing can be said about the arms of the angle. 1
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b) A plane divides the space into two regions (I) and (II) such that if a point A lies in region (I) and point B lies in region (II), then the straight line joining A and B will intersect the plane at one and only one point. If A and B lie in (I) or in (II) then the straight line joining A and B has no points in common with the plane. /(1) / (I) _P ___ Ill (Ill (Ill Figure 1.3 Figure 1.4 Figure 1.5 c) A plane P can be rotated about an axis x'x thus passing through all the points of space in one complete rotation A c I B I Figure 1.6 2
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1.4 Axioms a) Through three noncollinear points one and only one plane can be drawn. b) In every plane there are at least three noncollinear points. c) If two planes have one common point, then they have a second point in common. 1.5 Determination Of A Plane 1. 5 .1 Theorem Three noncollinear points A, B and C determine one and only one plane. Given three noncollinear points A, B and c. Prove that i) there exists a plane P passing through A, B and C; ii) P is the only plane that can contain A, B and C. B c A p Figure 1. 7 PROOF: i) and ii) can be proven by using axiom 1.4.a, and the proof is complete. I 1. 5. 2 Theorem A straight line and an external point determine one and only one plane. 3
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Given a straight line 0 and a point D. Prove that i) D and A determine a plane P; ii) P is the only plane that can be drawn through A and D. 1 c D a Figure 1.8 PROOF: i) Take any two points B and c on D. Then A, B and c are three noncollinear points which determine a plane P by Theorem 1.5.1. Since B and care two points in plane P, and at the same time they lie on D, then D lies completely in plane P by the definition of a plane. Hence P contains A and 0, or plane P is determined by A and D. ii) P is unique: Suppose there is another plane Q that contains A and D. Now, since D lies in P and Q, then B and C which lie on 0 must lie in both P and Q, then the three noncollinear points A, B and C are common to the two planes P and Q, which is impossible because three noncollinear points determine one and only one plane (Theorem 1.5.1). Hence, our supposition is wrong anQ plane P is the only plane that can be drawn through A, B and C. 4
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I 1. 5. 3 Theorem Two intersecting straight lines determine one and only one plane. Given two intersecting straight lines D and D' such that DnD' = A. Prove that i) D and D' determine a plane P; ii) P is unique. D D' Figure 1.9 PROOF: i) Take any two points B and C on D and D' respectively. Since A, B and C are three noncollinear points, they determine a plane P by Theorem 1.5.1. Since A and B are common to D and P, then D lies completely in P by the definition of a plane. Since A and C are common to D' and P, then D' lies completely in P. Hence D and D' both lie in plane P and determine P. ii) Suppose there is another plane Q that contains D and D'. Since D and D' are both in Q, then, B and C, which on D and D', lie in Q (definition of a plane). Hence the three noncollinear points A, B, and C are common to both 5
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planes P and Q which is impossible by Theorem 1.5.1. Thus our supposition is wrong and P is unique. I 1.5.4 Definition Two straight lines are said to be parallel if they lie in the same plane and dO not meet however they are extended. 1. 5. 5 Theorem Two parallel straight lines determine one and only one plane. Given two parallel straight lines D and D'. Prove that i) D and D' determine a plane P; ii) P is unique. B c 0' 1!:Figure 1.10 PROOF: i) Since D and D' are parallel straight lines, then they lie in a plane P. Hence D and D' determine a plane P (definition 1.5.4). ii) Suppose there is another plane Q determined by D and D'. Consider a point A on D and two points Band Con Since D and D' are both in P and Q, then A, B and C, which lie on D and D' must lie in P and Q, by the 6
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! I definition of a plane. But this is impossible, because three noncollinear points determine one and only one plane. Hence our supposition is wrong and P is unique. I 1.6 Conclusion A plane is determined by: a) three noncollinear points; b) A straight line and an external point; c) two intersecting straight lines; and d) two parallel straight lines. 1.7 Relative Positions Of Two Straight Lines Given two straight lines D and D', consider a point A on D'. The straight line D and the external point A determine a plane P. Hence the following discussion: i) If D' has another point B in common with P then D' will lie completely in P because D' has two points A and B in P (definition of a plane). In this case, D and D' are said to be coplanar. D D' A B p Figure 1.11 7
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ii) If D' cuts P at one and only one point A, then D and D' cannot lie in one plane. D I D' I Figure 1.12 PROOF: Suppose that D and D' lie in one plane Q. Now, since D' lies in Q then A lies in Q by the definition of a plane. Therefore, the straight line D and the external point A are common to both P and Q P and Q coincide* D', which lies in Q, must lie in Pa contradiction, since D' cuts P in one and only one point A. Hence, our supposition is wrong and D and D' cannot lie in one plane. D and D' are called skew straight lines. I Conclusion: i) Two straight lines are coplanar if they lie in the same plane; in this case they can be either intersecting or parallel. ii) Two straight lines are called skew if they do not in the same plane. 8
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1.8 Theorem I q If two distinct planes have a point in common, then they have a straight line in common. This straight line is the locus of points common to both planes. l Given two distinct planes P and Q, that have a common point A. Prove that i) P and Q have straight line t in common; ii} t is the locus of points common to P and Q. t p Figure 1.13 PROOF: i) Since P and Q have a point A in common, then they have a second point B in common by axiom 4.c. The straight line t joining A and B is the required straight line, i.e., t is common to both P and Q because t has two distinct points A and B which lie in both P and Q from the definition of a plane. ii.a) Every point c common to P and Q lies on t. PROOF: Suppose that c does not lie on t, then the three noncollinear points A, B and C are common to P and Q P and Q 9
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must coincide because three noncollinear points determine one and only one plane, and this is a contradiction since P and Q are distinct by hypothesis. Hence, our supposition is wrong and every point C common to P and Q lies on t. b) Conversely, every point M that lies on t is common to P and Q. PROOF: Since M lies on t and t lies in both P and Q, then M is common to P and Q from the definition of a plane. Therefore, from (a) and (b), we conclude that t is the locus of points common to P and Q. I 1.9 Definition Of A Tetrahedron A tetrahedron, also called a gauche quadrilateral or a skew quadrilateral, is a solid whose opposite edges are skew two by two. In the figure, ABCD is a tetrahedron where AC and BD are skew; AD and BC are skew; and AB and CD are skew. AC, BD, AD, BC, AB, and CD are called edges. A, B, c, and D are called the vertices of the tetrahedron. A B c Figure 1.14 10
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1. 9 .1 Problem Consider two planes P and Q that intersect along a straight line D. Also, consider two triangles ABC and A'B'C' that lie in P and Q respectively. A'A, B'B, and C'C are concurrent at a point V. AB, BC, and AC intersect A'B', B'C', and A'C' at M, N and R respectively, prove that M, N and R are collinear Figure 1.15 ..... 1/ /: '' / PROOF: This problem is an application of theorem 1.8. M lies on AB and A'B', and AB and A'B' lie in P and Q respectively. Therefore, M lies in both P and Q by the definition of a plane. Similarly R and N lie in both P and Q. Since M, N and R are common to P and Q, then M, N and R must lie on D which is the line of intersection of P and Q by theorem 1.8. But 0 is a straight line, so M, N and R are collinear because they lie on a straight line D : P n Q. I 11
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1.9.2 Problem Consider a tetrahedron ABCD where B, c and D are fixed while A moves along a fixed straight line J. Find the locus of the midpoint of AC. I I I D c Figure 1.16 Solution: The fixed straight line J and the external fixed point C determine a fixed plane P. Let M be the midpoint of AC. Draw J and J. Since C is fixed and J is fixed, CH = h = constant. Then, we know from plane geometry, MH' = (CH)/2 = h/2 = constant. Since J is fixed and M is moving such that its distance from J is constant, the locus of M is a fixed straight line t parallel to J and h/2 units away from J, and t lies in the fixed plane P. I p c Figure 1.17 12
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2. Straight Line Parallel To A Plane 2.1 Necessary And Sufficient condition For A Straight Line To Be Parallel To A Plane 2.1.1 Definition A straight line D is said to be parallel to a plane P if D and P have no points in common. 2.1.2 Theorem A necessary and sufficient condition for a straight line D, which does not lie in a plane P to be parallel to this plane is that D shall be parallel to a straight line in P. i) Necessary Condition: If a straight line D is parallel to a plane P, then D is parallel to a straight line lying in P. Given a straight line D that is parallel to plane P. Prove that D is parallel to a straight line D' that lies in P. p Figure 2.1 PROOF: Take any point A in P. The straight line D and the external point A determine a plane Q from theorem 1.5.2. planes P and Q have a point A in common, P and Q have a 13
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straight line D' in common by theorem We want to show that Dis parallel to D'. D and D' are coplanar in plane Q. It remains to show that D and D' have no point in common. To prove this, let us suppose that D and D' have a point M in common. Now Me D' and D'c P hence Me P But M = D n D' Me D. Since M belongs to D and P, D cuts P at M. But this is impossible because D is parallel to P, meaning that D and P have no points in common. Therefore, our supposition was wrong and D and D' have no points in common. Since D and D' have no points in common, and they are coplanar in plane Q, then D and D' are parallel, and D is parallel to a straight line D' that lies in P. ii) Sufficient Condition: If a straight line D, not lying in P, is parallel to a straight line D' in P, then D is parallel to P. Given a straight line D' in plane P and a straight line P such that D and D' are parallel. Prove that D and P are parallel. PROOF: Suppose that D is not parallel to P. We know from hypothesis that D does not lie in P. Therefore, D cuts P at M. Since D and D' are parallel, then they determine a plane Q that cuts P along D'. M is a point common toP and Q. M lies on D' = P n Q, by theorem 1.8. Therefore, D cuts D' at M, 14
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I I but this is impossible because, by hypothesis, D and D' are parallel. Hence, D cannot cut P and D and P are parallel. 0 Figure 2.2 I 2.2 Axiom Through any point in space, there is one and only one straight line parallel to a given straight line. 2.3 Theorem If a plane P intersects one of two parallel straight lines D and D', then it will intersect the other in one and only one point. Given two parallel straight lines D and D' and plane P cuts D at A. Prove that P cuts D' in one and only one point. p Figure 2.3 [) A : 0 0' B B' 15
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PROOF: Since D and D' are parallel, D and D' determine a plane Q from theorem 1.5.4. P and Q have a point A in common, so P and Q have a straight line t in common by theorem 1.8. Since t cuts D at A, then it should cut its parallel D' at B (from plane geometry). Since Bet and t = P n Q, then Be P. But BeD', therefore, P cuts D' at B. We want to show that B is unique. Suppose that P cuts D' at another point B'. Then A, B and B' will be three noncollinear points common to P and Q, soP and Q, by theorem 1.5.1, coincide. Therefore, D, which lies in Q, should lie in P. But this is a contradiction to the hypothesis that P cuts D at A. So, B is the only point of intersection between P and D', and P cuts D' at one and only one point B. 2.4 Theorem If two straight lines are each parallel to a third straight line, then they are parallel to each other. Given three straight lines ct1 d2 and d 3 such that d1 is parallel to d3 and d2 is parallel to d3 Prove that d1 is parallel to d2 16 I Figure 2.4
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I i I I i I I PROOF: Take any point A on d 2 Then d 1 and A determine a plane P by theorem 1.5.2. Since plane P has a point A in common with ct2 then plane P either cuts ct2 or contains i.t. Suppose that P cuts d 2 at A. Then, according to theorem 2.4, plane P will intersect d 3 which is parallel to d 2 If plane P cuts d 3 then P will cut d 1 that is parallel to d3 for the same reason. But this is impossible because d 1 lies in P. Therefore, we have reached a contradiction and plane P must contain d 2 Hence d 1 and d 2 are coplanar in plane P Now, suppose that d 1 and d 2 meet at a point M. Then, through one point M, we will be drawing two straight lines d 1 and d 2 parallel to d 3 which is impossible according to axiom 2.2. Hence ct1 and ct2 cannot meet; are coplanar; and, by the definition of two parallel straight lines, are parallel. I 2.5 Theorem Through one of two skew straight lines, one and only one plane can be drawn parallel to the other line. Given two skew straight lines o 1 and o2 Prove that through o 2 one and only one plane P can be drawn parallel to o 1 17
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Figure 2.5 PROOF: Take any point A on o 2 Through A draw a straight line d parallel to o 1 Since d and o 2 are intersecting they determine a plane P by theorem 1.5.3. o 1 is parallel to P because o 1 is parallel to a straight line d in P, which is a sufficient condition. Therefore, through o 2 we can draw a plane P parallel to o 1 P is unique: Suppose there is another plane Q passing through o 2 and parallel to o 1 Then through A e o 2 we can draw a straight line d' in Q parallel to o 1 which is a necessary condition. Then through A, we will be drawing two straight lines d and d' parallel to o 1 which is impossible according to axiom 2.2. Therefore, we have reached a contradiction and P is the only plane. I 2.6 Theorem If two intersecting planes P and Q pass through two straight lines 0 and 0', then the line of intersection xy of P and Q is parallel to D and 0'. 18
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Given two intersecting planes P and Q, xy = P n Q, 0 and D' are parallel, D c P, and D' c Q. Prove that the two parallel straight lines D and D' are parallel to xy. Figure 2.6 PROOF: Since D is parallel to a straight line D' in Q then Dis parallel to plane Q (sufficient condition). Since D and xy are coplanar in plane P, they are either parallel or intersecting. Suppose they intersect at a point A. Since xy c Q, A which lies on xy, should lie in Q. So, D cuts Q at A, which is impossible because we have just shown that D is parallel to Q. Therefore, D cannot cut xy. D and xy are coplanar in plane P; and they cannot meet, therefore, by definition of two parallel straight lines, D and xy are parallel. Similarly, D' is parallel to xy. Hence, by theorem 2.4, the two parallel straight lines D and D' are parallel to xy. 19
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2.7 Theorem If a straight line D is parallel to two intersecting planes P and Q, then it is parallel to their line of intersection. Given two planes P and Q; xy = P n Q; D is parallel to P; and D is parallel to Q, prove that 0 is parallel to xy. 0 Figure 2.7 PROOF: Since 0 is parallel to P and 0 is parallel to Q, 0 is parallel to o 1 in P and D is parallel to o 2 in Q (Necessary condition). According to theorem 2.4, o 1 is parallel to o 2 Now, o 1 is parallel to o 2 o 1 c P and o 2 c Q so by theorem 2.6, the parallel lines n1 and n2 are parallel to xy. Therefore, by theorem 2.4, D is parallel to xy. N.B.: Actually, D, o 1 D 2 and xy are parallel straight lines. I 2.8 Theorem If three planes P,Q, and R intersect two by two then the lines of intersection are either parallel or concurrent. 20
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Give three intersecting planes P, Q, and R with o 1 = PnR, o2 = QnR, and o 3 = PnQ, prove that o 1 o2 and o 3 are either parallel or concurrent. Figure 2.8 PROOF: Suppose that o 1 and o 2 intersect at M. Then M will be a point that belongs to P and Q because M lies on o 1 and o 2 that lie in P and Q. Now, since M lies in P and Q then M will lie on n 3 which is the intersection of P and Q, by theorem 1.8. Hence o 1 o 2 and o 3 are concurrent at M. Now, suppose that o 1 and o 2 are parallel. Then p and Q will contain two Figure 2.9 straight lines n 1 and o 2 respectively that are parallel to each other. Then 0 1 o2 and o3 are parallel by theorem 2.6. I 21
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2.9 Angle Between Two Skew Straight Lines Consider two skew straight lines D and D'. Take any point A on D'. The angle xAy, as shown in the figure, obtained by drawing Ax parallel to D, is called the angle between D and D'. If the angle between D and D' is 90, then D and D' are called orthogonal straight lines. D D' Figure 2.10 22
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3. Parallel Planes 3.1 Necessary And Sufficient Condition For Two Planes To Be Parallel 3.1.1 Definition Two planes P and Q are said to be parallel if they have no points in common. 3.1.2 Theorem A necessary and sufficient condition for two planes P and Q to be parallel is that one of them contains two intersecting straight lines parallel to the other plane. 3.1.3 Necessary Condition If two planes P and Q are parallel, then one of them contains two intersecting straight lines parallel to the other plane. Given two planes P and Q such that P and Q are parallel, prove that plane P contains two intersecting straight lines parallel to Q. X A Figure 3.1 23
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PROOF: Take any point A in plane P. Through A draw any two straight lines Ax and Ay in plane P. We want to show that Ax and Ay are parallel to plane Q. Suppose that Ax is not parallel to Q. Then Ax will intersect plane Q at M. Since Axe P, Me P, and M is common to P and Q. But this is impossible because P and Q can have no points in common since they are parallel (by hypothesis). Therefore, our supposition is wrong and Ax is parallel to Q. Similarly, Ay is parallel to Q. So, plane P contains two intersecting straight lines Ax and Ay parallel to plane Q. N.B.: Since A is arbitrary, any two intersecting straight lines in plane P are parallel to plane Q. Following the same proof above, we can prove that every straight line in plane P is parallel to plane Q. 3.1.4 Sufficient Condition If two intersecting straight lines in plane P are parallel to plane Q, then P and Q are parallel to each other. Given two planes P and Q, and plane P contains two intersecting straight lines Ax and Ay parallel to Q, prove that P and Q are parallel. PROOF: Suppose that P and Q are not parallel, then P and Q must intersect along a straight line t. Now Ax and t are coplanar in plane P, and cannot meet because if Ax meets which lies in Q, then Ax will be intersecting plane Q. But this is impossible since Ax is parallel to Q. Therefore, Ax is 24
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parallel to t, by definition of two parallel straight lines. By a similar argument, we can prove that Ay is parallel to t. Now, Ax and Ay are intersecting, and we are drawing two intersecting straight lines, in the same plane, parallel to a third straight line, which is impossible (by plane geometry). Therefore, our supposition is wrong and P is parallel to Q. Figure 3.2 3.2 Theorem Through a given point, not lying in a plane P, one and only one plane Q can be drawn parallel to P. Given a plane P and a point P. Prove that, through A, one and only one plane Q can be drawn parallel to P. PROOF: i) Construction of Q: Take any point A' in plane P. Through A' draw any two straight lines A'x' and A'y'. Through A draw Ax and Ay parallel to A'x' and A'y' 25
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I I I respectively. Since Ax and Ay are intersecting, they determine a plane Q (theorem 1.5.3) which is the required I plane. Figure 3.3 ii) Proof of the construction: Since Ax is parallel to a straight line A'x' in plane P, Ax is parallel to P (sufficient condition for a straight line to be parallel to a plane). Similarly, Ay is parallel toP. Therefore, plane Q, determined by Ax and Ay, is parallel to plane P (sufficient condition for two planes to be parallel). iii) Q is unique: Suppose there is another plane Q' through A parallel to plane P. Since plane Q' cuts Ax at A, it must cut its parallel A'x' (if a plane cuts one of two parallel straight lines, then it cuts the other in and only one point). If Q' cuts A'x' at some point N, then N, which is a point of A'x'c P, will be a point of 26
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P. Hence P and Q' have a point N in common which is impossible because Q' is supposed to be parallel to P. Therefore, we have a contradiction and Q is unique. I 3.3 Theorem If a plane intersects one of two parallel planes, it also intersects the other plane, and the lines of intersection are parallel. Given two parallel planes P and Q, and A plane R that intersects P along t, prove that R intersects Q along t', and t is parallel tot'. Figure 3.4 27
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PROOF: Suppose that R does not intersect Q. Then R will be parallel to Q. Take any point A on t. Then through A we will be drawing two planes P and R parallel to Q which is impossible because through a given point outside a given plane, one and only one plane can be drawn parallel to the given plane. Hence our supposition is wrong and R cuts Q along t'. Now, t and t' are coplanar in plane R, and they cannot have a point in common because they lie in parallel planes. Therefore, t and t' are parallel. I 3.4 Theorem If a straight line cuts one of two parallel planes, then it cuts the other. Given two parallel planes P and Q, and t is a straight line that cuts P at A. Prove that t cuts Q. 0 Figure 3.5 28
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PROOF: Suppose that t does not cut Q. Then t will be parallel to Q. If t is parallel to Q, then it will be parallel to a straight line t' in Q (necessary condition for a straight line to be parallel to a plane). Since plane P cuts tat A, it must cut its parallel t' {theorem 2.3). If P cuts t' at M, then M will be a point common to P and Q. But this is impossible because P and Q are parallel. Hence t cuts Q. 3.5 Theorem I The locus of straight lines drawn through a given point A parallel to a given plane P is a plane Q through A parallel to P. Given a plane P and a point P. Prove that the locus of all straight lines through A parallel to P is a plane Q parallel to P. y Figure 3.6 29
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PROOF: Through A draw a plane Q parallel toP (theorem 3.2). Then i) Every straight line Ax drawn through A parallel to P lies in Q since, if Ax does not lie in Q, then it will cut Q at A. Consequently, it will cut P which is parallel to Q by theorem 3.3. But this is impossible because Ax is parallel to P. Hence we have reached a contradiction and Ax lies in Q. ii) Since Q is parallel to P then every straight line in Q is parallel toP (3.1). Then Ay which lies in Q is parallel to P. Hence, every straight line Ay drawn through A in plane Q is parallel to P. Combining (i) and (ii), we deduce that the locus of straight lines drawn through A parallel to P is the plane Q drawn through A parallel to P. I 3.6 Theorem If the arms of two angles in space are parallel and in the same sense two by two, then the two angels are equal. If two arms are parallel and in the same sense, while the other two arms are parallel and in opposite sense then the two angels are supplementary. i) Given two angles xAy and x'A'y' in space such that Ax and A'x' are parallel, Ay and A'y' are parallel, Ax and A'x' 30
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are in the same sense, and Ay and A'y' are in the same sense. Prove that PROOF: Take points C,B,C', and B' on Ax, Ay, A'x', and A'y' respectively such that AC=A'C',and AB=A'B'. Since AC and A'C' are two parallel straight lines, they determine a plane ACC'A' (two parallel straight lines determine a plane). Since the vectors AC=A'C', then ACC'A' is a parallelogram (plane geometry). Since ACC'A' is a parallelogram, then vectors s A'A=C'C (1). By a similar Figure 3.7 argument to that used above, we prove that ABB'A' is a parallelogram which implies that the vectors A'A=B'B (2). Comparing (1) and (2), we get vectors C'C=B'B. That means C'C and B'B are equal in magnitude, parallel and have the same sense (to verify why C'C and B'B are parallel, we can say that C'C and B'B are both parallel to A'A, which implies that B'B and C'C are parallel). Since C'C and B'B are parallel, CBB'C' is a plane. And since the vectors C'C=B'B, CBB'C' is a parallelogram, which implies that CB=C'B' (plane geometry). Consider the two triangles ABC and A'B'C'. Then AC=A'C', AB=A'B', and CB=C'B'. Therefore, AABC:AA'B'C' (plane geometry), and angles xAy and 31
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x'A'y' are equal (corresponding parts of congruent triangles are equal). ii) Given that the lines Ax and A'x' are parallel and have the same sense; Ay and A'y' are parallel and have opposite senses, prove that angles xAy + x'A'y' = 180'. PROOF: Produce yA toy'', Then we will have two angles Ax'A'y' and AxAy'' as in case (i), which implies that LxAy''= lx'A'y'. But lxAy + lxAy'' = 180'. Therefore, lxAy + lx'A'y' = 180' (by substitution). I 't (,(,"''I A I A" "L';JG'. ))))'f. Figure 3.8 3.7 Theorem Two planes P and Q parallel to a third plane R, are parallel to each other. Given three planes P,Q, and R such that P is parallel to R and Q is parallel to R. Prove that P is parallel to Q. 32
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I Figure 3.9 PROOF: Suppose that P and Q are not parallel, then they will intersect along a straight line t. Now through a point Ae t, we will have two planes P and Q parallel to the same plane R. But this is impossible by theorem 3.2. Therefore, P and Q are parallel. I 3.8 Theorem Two parallel planes cut off equal segments on two parallel straight linea. Given two parallel planes P and Q, a straight line D cutting P and Q at A and B respectively, and a straight line D' parallel to D cutting P and Q at A' and B' respectively, prove that AB=A'B'. 33
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p B+.e Figure 3.10 PROOF: The two parallel straight lines D and D' determine a plane R. Plane R cuts the parallel planes P and Q along two parallel straight lines AA' parallel to BB' (theorem 3.3). The quadrilateral ABB'A' that has its opposite sides parallel is a parallelogram. Therefore, its opposite sides are equal and AB=A'B'. I 3.9 Theorem Three or more parallel planes cut off proportional segments on any two transversals. Given two transversals D and D'; and three parallel planes P, Q and R cutting DatA, Band c, and D' at A', B' and C' respectively, prove that (3.1) 34
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0 0' A' a C" \ Figure 3.11 PROOF: The parallel through A to D' cuts the planes Q and R at M and N respectively. The straight lines BM and CN are parallel as the intersections of two parallel planes by a third. Hence from plane geometry, we can write (3.2) But AM= A 1 B 1 and MN = B 1 C 1 (Theorem 3.8). Substituting this into equation (3.2), we get (3. 3) which is the required. I 35
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l I 1 I I I 3.10 Converse If three straight lines cut proportional segments on two transversals in the same sense, then three parallel planes can be drawn through these three straight lines. Given three straight lines AA', BB' and CC' cutting proportional segments on two skew straight lines D and D' in the same sense, that is Prove that three parallel planes P, Q and R can be drawn through AA', BB' and CC'. D Figure 3.12 D' 36 c c (3.4)
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PROOF: Draw Ax parallel to BB' and By parallel to AA'. The planes P(AA',Ax) and Q(BB',By) are parallel because two intersecting straight lines in one are parallel to two intersecting straight lines in the other. Let R be the plane drawn through c parallel toP and Q. Suppose R cuts D' in C'' Then according to theorem 3.8, we can write (3.5) but (3. 6) Comparing relations (3.5) and (3.6), we get (3. 7) which shows that C'' coincides with C' and consequently C' lies in R. Therefore, three parallel planes P, Q, and R can be drawn through AA', BB' and CC'. I 3.11 Some Problems On Chapters Two And Three 1. Consider a tetrahedron ABCD in which AB=CD=a. Let M be a variable point on BC such that BM=x. The parallels through M to AB and CD cut AC and BD in R and P respectively. The plane MPR cuts AD in Q. a) Show that the quadrilateral MPQR is a parrn. 37
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b) Show that, as M traces BC, the perimeter of MPQR remains constant. PROOF: A ' ' _::::..... pD c Figure 3.13 a) CD is parallel to the plane MPQR because it is parallel to a straight line MP in MPQR. Hence CD and RQ have no points in common; but CD and RQ are coplanar in plane ACD, thus CD is parallel to RQ. But MP is parallel to CD. so, CD is parallel to RQ and CD is parallel to MP, which imply that MP is parallel to RQ (if two straight lines are parallel to a third straight line, then they are parallel to each other). Similarly, MR is parallel to PQ. Therefore, MPQR is a parallelogram. b) is similar to so MR BA CM_MR CB a CM MR CM+ X 38 a(CM) CM +X (3.8)
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is similar to ABCD, so MP CD BM MP BC a X MP= ax x+CM x+CM (3.9) perimeter of MPQR 2MR + 2MP = Za(CM) + CM +X 2a(CM+x) (CM + x) 2ax =2a(CM)+2ax x+CM CM+x 2a = constant (3.10) I 2. A fixed circle C of center 0 and radius R lies in a plane P. Consider a fixed point V not lying in P and a variable point M on the circumference of C. The straight line VM cuts a fixed plane Q parallel to P in a point N. Find the locus of the point N as M traces c. v c Figure 3.14 39
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PROOF: The straight lines VM and VO cut plane P, which implies that VM and VO cut the parallel plane Q at N and 0' respectively. So, OM is parallel to O'N (intersections of two parallel planes by a third). aVOM is similar to aVO'N o 1 N vo' o'N vo' = = OM VO R VO o1N = R vo' = constant vo (3.11) So the locus of N is a circle, centered at 0' with radius of vo' R that lies in the plane Q. VO I 3. Consider a plane P and two noncoplanar straight lines D and A cutting P at A and B respectively. Let M be a variable point in D. The parallel to A through M cuts P at a point M'. a) Find the locus of M' as M traces D. b) The plane P' drawn through M parallel to P cuts A at a point N. What is the nature of the quadrilateral MM'BN? c) Construct M when the distance MN has a given length 1. Discuss. 40
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A 0 Figure 3.15 PROOF: a) D is fixed and MM' has a fixed direction which is that of A. Therefore, the plane MM'A determined by D and the fixed direction of is a fixed plane. So d, which is the intersection of the two fixed planes P and MM'A, is a fixed straight line and is the locus of M'. b) Since MM' is parallel to NB, the parallel planes P' and P cut off equal segments MM' and NB on the parallel straight lines MM' and A. Since the vectors MM'BN is a parallelogram. c) Suppose the problem is solved, then we get M'B=l. Hence, to determine M it is enough to determine M' and then, through M', draw a straight line parallel to 4 to get M. so, let us find M'. M' is the intersection of a circle centered at'B with radius 1 and d=Pn(MM'A). 41
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i i I DISCUSSION: 1) If BM' = l>x = the distance from B to d, then there are two points M' and two points 2) If BM' = 1 = x, then there is one point M' and there is one point M. 3) If BM' ; l
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4. Perpendicular Straight Lines And Planes 4.1 Necessary And Sufficient COndition For A Straight Line To Be Perpendicular To A Plane 4.1.1 Definition A straight line D is perpendicular to a plane P if it is orthogonal to every straight line in that plane. 4.1.2 Theorem A necessary and sufficient condition for a straight line D to be perpendicular to P is that D should be orthogonal to two intersecting lines in P. 4.1.3 Necessary COndition If a straight line D is perpendicular to a plane P, then it is orthogonal to two intersecting straight lines in P. Given D perpendicular to P, and o 1 and o 2 are two intersecting straight lines in P. Prove that D is orthogonal 0 Figure 4.1 43
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PROOF: Since 0 is perpendicular to P 0 is orthogonal to all straight lines in P (definition 4.1) and, in particular, to two intersecting straight lines in P such as 01 and 02. 4.1.4 Sufficient Condition If a straight line 0 is orthogonal to two intersecting straight lines in P, then it is perpendicular to P. Given a plane P, two intersecting straight lines o 1 and 02 in P, and a straight line 0 which is orthogonal to 01 and 02. Prove that 0 is perpendicular to P. D Figure 4.2 PROOF: D cannot lie in P, otherwise we will have two intersecting straight lines perpendicular to a third in the same plane which is impossible. Moreover, D cannot be parallel to P, otherwise its parallels in P would be perpendicular to two intersecting straight lines in P, which is impossible. Therefore, 0 cuts P at a point 0. Let o 3 be any straight line 44
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in P. Draw through 0, the straight lines Ox, Oy and Oz parallel to D 1 D 2 and D 3 respectively. Then (4.1) and (4.2) Take on D, two equal segments OA = OA', and draw in P any secant L cutting Ox in B, Oy in c, and Oz in M. Each of the straight lines Ox and Oy is a perpendicular bisector of AA' BA = BA' and CA = CA'. The two triangles ABC and A'BC having their sides respectively equal are congruent the angels ABC and A'BC are equal. Now consider the triangles ABM and A'BM. They have BM common, BA=BA' and the angles ABM = A'BM. Therefore, these triangles are congruent AM = AM' and the triangle AMA' is isosceles. The straight line OM, which is a median from the vertex of the isosceles triangle AMA', is also a height. Hence D is perpendicular to Oz and, consequently, it is orthogonal to its parallel D 3 which is any straight line in P. Therefore, D is orthogonal to every straight line in P, that is Di P (by definition 4.1.1). 4.2 Theorem Of Three Perpendiculars If, from an external point A, the perpendicular AO to i plane P and the perpendicular AH to a straight line De P are drawn, then OH is perpendicular to D. 45 I
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Given AO P, De P and AH D. Prove that OH L D. A D p Figure 4.3 PROOF: AO P AO D c P. Now, D i AO and D L AH 0 L plane AOH because D is perpendicular to two intersecting straight lines AO and AH in plane AOH. Now, D L plane AOH D OH c AOH. I 4.3 Theorem If two straight lines D and D' are parallel, every plane P perpendicular to one of them is perpendicular to the other. Given D I D' and D P. Prove that D' P. D D' A A' Figure 4.4 46
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PROOF: Since 0 L P 0 is orthogonal to all straight lines lying in P. Therefore, its parallel 0' is orthogonal to all straight lines lying in P 0' L P. I 4.4 Theorem If two planes P and Q are parallel, every straight line 0 which is perpendicular to one of them is perpendicular to the other. Given P I Q and 0 L P. Prove that 0 L Q. J o, >< o, o; 0 ><; 0 Figure 4.5 PROOF: Consider any two intersecting straight lines o1 and o2 in P. We have P I Q o 1 I Q and o 2 I Q. Hence, o 1 and o 2 are respectively parallel to two intersecting straight lines o 1 and o 2 in Q. We have 0 L P 0 L o 1 and 0 L o 2 Therefore, I 47
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4.5 Theorem Through a given point o one and only one Plane P can be drawn perpendicular to a given straight line o. I 1 Case 1 : Oe D I 'I a) Existence: consider two planes Q and R passing through D. Let Oxc Q and Oyc R such that Ox i D and Oy i D. The plane P determined by Ox and Oy is a plane perpendicular to D. 0 Figure 4.6 b) P is unique: Suppose P' is another plane perpendicular to D at 0. P' cannot contain Ox and Oy at the same time otherwise it coincides with P. So suppose that P' does not contain Ox. Therefore, P' and Q are distinct because P' D c Q. P' and Q are not parallel because they have a point 0 in common. Hence P' and Q intersect along a straight line Ox'. We have: DiP' D i Ox' c Q and DiP D i OX c Q. Consequently, we have two intersecting straight lines and Ox' perpendicular to a third straight line D in the same 48
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plane, which is impossible. Therefore, P' cannot be perpendicular to D at o, so P is the only plane. Case 2: Of D. a) Existence: Let D' be the parallel drawn through 0 to D and P the plane perpendicular to D' at 0 (case 1). P, being perpendicular to D', is perpendicular to its parallel D. D D' Figure 4.7 b) P is unique: If P' is another plane passing through 0 and perpendicular to D, P' would be perpendicular to D' which is parallel to D. But this is impossible because P is the only plane perpendicular to D' at o. Therefore, P is the only plane. 4.6 Corollary Two planes P and P' perpendicular to the same straight line D are parallel. Given D P and D P'. Prove that PI P'. 49
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D 0 A d Figure 4.8 PROOF: Suppose that P and P' are not parallel. Then they will be intersecting along a straight line t. Take any point Ae t. Then, through A, we will be drawing two planes P and P' perpendicular to D, which is impossible by theorem 4.5. Therefore, Pi P'. 4.7 Theorem Through a given point o, one and only one straight line D can be drawn perpendicular to a given plane P. Case 1: Oe P a) Existence: Let Ox be any straight line in P, Q the plane perpendicular to ox at o and Oy=PnQ. Draw, in Q, the straight line OA L Oy. We have: Ox L Q Ox L OA c Q. 50 I
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Now, OA Oy and OA Ox OA P (sufficient condition for a straight line to a plane). y X Figure 4.9 b) OA is unique: Suppose OA' is another perpendicular to P at 0. OA and OA' determine a plane R. Let Oz=PnR. We have OA P, OA' P and Oz c P OA Oz and OA' Oz which is impossible because OA, OA' and Oz are coplanare Consequently, OA is the only to P at 0. A z Figure 4.10 51
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, I Case 2: P a) Existence: Let P' be the plane passing through 0 and parallel to P and D the perpendicular to P' at o. We have: P I P' and D i P' D i P. < Figure 4.11 ; 0 f D' b) D is unique: If D' is another perpendicular to plane P through 0, D' would be perpendicular to P' which is parallel to P. But this is impossible because D is the only perpendicular to P' at 0. Therefore D is unique. I 4.8 Corollary Two straight lines D and D' perpendicular to the same plane P are parallel. Given DiP and D' i P. Prove that D I D'. 52
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D D' 0 Figure 4.12 PROOF: Let O=DnP and O'=D'nP. Suppose that D and D' are not parallel and let D'' be the parallel drawn through 0' to D. We have: D'' I D and D L P D'' L P. Consequently, D' and D'' are perpendicular to P at the same point 0' which is impossible. Therefore, D'' coincides with D', and the two straight lines D and D' are parallel. I 4.9 Theorem The set of straight lines, passing through a given point 0 and which are orthogonal to a given straight line D, is the plane passing through 0 and perpendicular to D. Given a straight line D and a point 0. Prove that the locus of all straight lines passing through 0 and perpendicular to D is a plane perpendicular to o. PROOF: a) Every straight line Oxc Pis orthogonal to D a line perpendicular to a plane). 53 I ... I
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D D X Figure 4.13 Figure 4.14 b) Every straight line Oy passing through 0 and perpendicular to D should lie in P, otherwise Oy and any straight line Oxc P determine a plane P' perpendicular to D and passing through o. But this is impossible because P is the only plane passing through 0 and perpendicular to P. Therefore, P' coincides with P and, consequently, Oy lies in P. Conclusion: Hence, plane P is the locus of straight lines drawn through o and orthogonal to D. This is true if Oe D or D, and the proof is the same for both cases. I 4.10 Theorem If a straight line D is perpendicular to a plane P, every straight line D' which is orthogonal to D and not lying in P is parallel to P. Given D L P and D' L D with P. Prove that D' I P. 54
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D o Figure 4.15 PROOF: If 0' has a point 0 in P, then it would lie in P (theorem 4.9). But this is impossible because P. Therefore, 0' is parallel to P. I 4.11 Converse If a straight line D' is parallel to a plane P, then every straight line 0 perpendicular toP is orthogonal to 0'. Given 0 L P and 0' I P. Prove that 0 L 0'. D o D .. Figure 4.16 PROOF: Since 0' I P 0' 0' c: P (if a straight line is parallel to a plane, then it is parallel to a straight line 55
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lying in the plane). We have: DiP DiD'' c P Dis orthogonal to D' which is parallel to D''. I 4.12 Perpendicular and Obliques 4.12.1 Theorem If, from a given point not lying in a given plane, the perpendicular and many obliques are drawn to this plane then a) the perpendicular is shorter than any oblique b) Two obliques meeting the plane at points equidistant from the foot of the perpendicular are equal c) Two obliques meeting the plane at points unequally distant from the foot of the perpendicular are unequal and the one cutting the plane at the more remote point is the greater. PROOF: a) Given A P, AO i P, and AB is any oblique. Prove that AO < AB. Figure 4.17 PROOF: AO i P AO i OB c P. Therefore, the triangle AOB is a right triangle and hence its leg is less than its hypotenuse AB. 56
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b) Given A f P, AO P, and AB and AC are obliques such that OB=OC. Prove that AB=AC. A B c Figure 4.18 PROOF: Consider the two triangles AOB and AOC. AO P AO OB c P and OA OC c P. The side AO is common to these triangles and the sides of OB and OC are equal. Therefore, these triangles are congruent and AB=AC, as required. c) Given A P, AO P, and AB and AC are two obliques such that OBAB. A B c Figure 4.19 PROOF: BO c P and AO OC c P. Apply Pythagoras' theorem on the two right triangles AOB and AOC. 57 Then
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and By hypothesis, OBAB. (4.3) (4.4) (4.5) I If a point, not lying in a given plane P is joined to every point in this plane, then a) The shortest one of these segments is perpendicular to P b) Two equal obliques meet the plane at points which are equidistant from the foot of the perpendicular c) Two unequal obliques cut the plane at points unequally distant from the foot of the perpendicular and the greater meets the plane at the more remote point. PROOF: a) Given A P, and AO is the shortest one of the segments drawn from A to every point A .!\. in P. Prove that AO P. Figure 4.20 58
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PROOF: Suppose that AO is not perpendicular to P. Draw AO' P. Then, according to theorem 4.12, AO'AB. Prove that OC>OB. A c Figure 4.22 59
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PROOF: The segments OC and OB cannot be equal otherwise AC and AB will be equal. Moreover, OB cannot be greater than OC otherwise AB will be greater than ACa In the above two cases, we reach a contradiction and the only possible remaining case is OC>OB. I 4.14 The Distance From A Point To A Plane The distance from a point A to a plane P, is the length of the segment of the perpendicular drawn from A to P and which is included between A and P. A J! fdistance Figure 4.23 4.15 Symmetry With Respect To A Point Two points A and A' are symmetrical with respect to a point 0 if 0 is the midpoint of the line segment AA'. A 0 A" Figure 4.24 A point 0 is said to be the symmetry center of a figure F if the symmetric of any point in F with respect to 0 belongs to F. 60
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4.16 symmetry With Respect To A straight Line Two points A and A' are symmetrical with respect to a straight line D if D is the perpendicular bisector of the line segment AA'. A straight line D is said to be an axis of symmetry of a figure F if the symmetric of any point in F with respect to D belongs to F. 4.17 symmetry with Respect To A Plane Two points A and A' are D A Figure 4.25 symmetrical with respect to a plane P if P is perpendicular to the line segment AA' at its midpoint. This plane is called the mediator plane of the line segment AA'. A plane Pis said to be a plane of symmetry of a figure F if the symmetric of any point in F with respect to P belongs to F. 4.18 Theorem The set of points which are equidistant from two given points A and B is the mediator plane of the segment AB. PROOF: Consider two fixed points A and B. Let o be the midpoint of the segment AB. P the mediator plane of AB and M any point in the space. we have MA=MB the aMAB is an isosceles triangle MO AB MO c P. Therefore, the points which are equidistant from A and B is the plane P. 61
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A 0 B Figure 4.26 4.19 Axis Of A Circle Consider, in a plane P, a circle C of center 0. The perpendicular at o to P is called the axis of circle C. D ,o c Figure 4.27 4.20 Theorem The set of points which are equidistant from three noncollinear points A,B, and C is the axis of the circumcircle of the AABC. 62
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D M A c B Figure 4.28 PROOF: Consider three noncollinear points A,B, and c. Let 0 be the circumcenter of D the axis of the circumcircle of the AABC, M any point in the space, and H the foot of the perpendicular from M to the plane ABC. we have MA=MB=MC HA=HB=HC H coincides with o M E D. Therefore, D is the set of points which are equidistant from the three points A,B and c. I 4.21 Some Problems On Chapter Four 1. Given a plane P and two points A and B external to P, find the set of the points Me P such that a) M.l\=MB b) ID\2 1m2 = k 2 where k is a constant. 63
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SOLUTION: I I I p Figure 4.29 a) MA=MB M e Q where Q is the mediator plane of AB. So, M e P and M e Q Me d=P0Q. Therefore, the set of M is d, the line of intersection of P and Q 0 Figure 4.30 b) Let 0 be the midpoint of AB, and let H be the foot of the L from M to AB. Then 64
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'Rli.' mf' = 21:m : a nd MER _, MEt=PnR. Therefore, the set of M is t, the line of intersection of P and R. I 2. Given a plane P, a fixed point A ffP and a variable point M EP such that AM=d (d being a given length). Find the set of the points M. A Figure 4.31 PROOF: From A draw a perpendicular to plane P. Call the foot o. 0 is fixed. A is fixed and 0 is fixed AO= 1 =constant. 65
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AO ..L P .. AO ..L OM c: P => .c.AOM is a right A. Apply Pythagoras' theorem on AAOM: ]jif' = ?511' + d2 = ?511' + 1. ?511' = d2 12OM= /d2 1" =constant (4.7) Therefore, the locus of M is a circle, with its center at 0 and radius = 1 2 and this circle lies in the fixed plane P. 3. Consider a plane P, a fixed point 0 E"P, a fixed point A $P and a variable point M in P. Find the set of the points M in each one of the following cases: a) The .!DAM = 900 b) The MOM = 900 c) The .l.AMO = 900 SOLUTION: a) OA is fixed and MA ..LOA at a fixed point A that the set of MA is a plane Q perpendicular to OA at A. Therefore, the set of M is d=PnQ, the line of intersection of the two plane P and Q. A Figure 4.32 66
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b) Draw AO' ..LP. 0' is fixed. AO' ..LP AO' ..LOM c::P. OM ..L OA and OM ..LAO' OM ..Lplane AOO' because OM is orthogonal I I to two intersecting straight lines in plane AOO'. OM Lto plane AOO' OM is ..L to 00' M lies on d which is OM produced because d is perpendicular to a fixed straight line 00' at a fixed point o. So d is a fixed straight line. A d Figure 4.33 c) Draw AO' ...LP. 0' is fixed. AO' ...LP AO' ..LOM c::P. Now, OM ..LMA and OM ..LAO' OM ..Lplane AO'M because OM is orthogonal to two intersecting straight lines in the plane AO'M. Now, OM ..LAO'M OM ..J...MO' c::p i10M0'=9QO. Since 6DM0';9QO, and 0 and 0' are fixed, then the locus of M is a circle whose diameter is 00', and this circle lies in plane P. Figure 4.34 I 67
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5. Perpendicular Planes 5.1 Definitions 5.1.1. Definition A dihedral angle (or dihedral) is the figure formed by two intersecting semiplanes limited by their line of intersection. The two semiplanes are called the faces of the dihedral and their line of intersection is the edge of the dihedral. Figure 5.1 In general, a dihedral angle is designated by its edge (xy), by its two faces and its edge (PxyQ), or simply by its two faces (P,Q). 5.1.2 Definition Two dihedral angles are said to be adjacent if they have the same edge and a common face between them. 5.1.3 Definition Two dihedral angles are said to be opposite by the edge (or R Figure 5.2 opposite) if the faces of one are the extensions of the faces of the other. 68
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Figure 5.3 5.1.4 Definition Two dihedral angles are said to be equal if they are superposable, that is if they can be made to coincide. 5.2 The Plane Angle Of A Dihedral The plane angle of a dihedral is the angle obtained by drawing through any point of the edge two semistraight lines perpendicular to this edge and each one of them lying in one face of the dihedral. If the plane angle is a right angle, then the dihedral is called a right dihedral. 69 Figure 5.4
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5.3 Necessary And Sufficient Condition For Two Planes To Be Perpendicular 5.3.1 Definition Two planes are perpendicular to each other if one of the four dihedral angles they form is a right dihedral. Figure 5.5 5.3.2 Theorem A necessary and sufficient condition for two planes to be perpendicular is that one of them contains a straight line perpendicular to the other. 5.3.3 Necessary Condition If two planes are perpendicular, then one of them contains a straight line perpendicular to the other. Given P Q. Prove that one of P and Q contains a straight line is perpendicular to the other. 70
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Figure 5.6 PROOF: Through any point 0 of the line of intersection of P and Q, draw OA and OB in planes P and Q respectively, perpendicular to xy. The &AOB, thus constructed, is one of the plane angles of the two planes P and Q. P Q is a right angle (definition of two perpendicular planes). Now, OA OB c Q and OA xy c Q OA Q. Similarly, OB P. 5.3.4 Sufficient Condition If a straight line is perpendicular to a plane Q, then any plane P passing through this straight line is perpendicular to Q. Given OA Q. Prove that any plane P passing through OA is perpendicular to Q. 71
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X Figure 5.7 PROOF: Any plane P passing through OA cuts the plane Q along a straight line xy that is in Q. We have OA L Q OA L xy c Q. Draw in Q, the perpendicular OB to xy, then the angle AOB is the plane angle of the dihedral (P,Q). But OA L Q OA L OB c Q6..AOB = 900 Therefore, the two planes P and Q are perpendicular. 5.4 Theorem (5.1) I If two planes are perpendicular, every straight line drawn in one of them perpendicular to their line of intersection is perpendicular to the other. Given P L Q, AO in P, and AO L xy. Prove that AO L Q. 72
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y )( Figure 5.8 PROOF: Draw, in Q, the straight line OB perpendicular to xy. The angle AOB is the plane angle of the dihedral (P,Q). Hence P J. Qfl.AOB = 9Qo AO J. OB (5.2) Since OA OB which is in Q and OA xy which is also in Q, then AO J. Q as required. 5.5 Theorem I If two planes are perpendicular and if from a point in one of them a straight line is drawn perpendicular to the other, then this straight line lies completely in the first plane. Given P J. Q, xy=PnQ and Az J. Q where A is in P. Prove that Az is in P. 73
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Figure 5.9 PROOF: From AE P draw AO i xy. Then AO is in P and AO i Q (theorem 5.4). Hence, if from Ae P, we draw Az i Q, then Az should coincide with AO because from a given point A one and only one perpendicular to a given plane Q can be drawn. consequently, Az is in the plane P. 5.6 Theorem If each of two intersecting planes P and Q is perpendicular to a third plane R, then their line of intersection xy is perpendicular to R. Given PnQ=xy. P i R and Q i R. Prove that xy i R. X Figure 5.10 74
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PROOF: From any point Ae xy, draw Az perpendicular to R. Now we have AE.xyAEPAzcp (5.3) from theorem 5.5. Similarly AExyAEQAz<=Q (5.4) Hence, Az=PnQ which implies Az coincides with xy. Consequently, xy R. I 5. 7 Theorem Through a given straight line D, not perpendicular o a plane P, one and only one plane Q can be drawn perpendicular to P. / 1". /:"\ Figure 5.11 PROOF: a) Existence: Through any point A on 0, draw the perpendicular AA' to plane P. The plane Q determined by AA' and 0 plane P because it contains a straight line AA' P. b) Q is unique: Every other plane Q' passing through 0 and is P should contain AA' by theorem 5.5. Consequently, Q' coincides with Q because these twa planes have two intersecting straight lines D and AA' in common. 75
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If Dis in P, the same construction and the same proof remain valid. 5.8 Theorem I If a straight line D and a plane Q are both perpendicular to the same plane P, then D is either parallel to Q or lies in it. Given D P and Q i P. Prove that D is parallel to Q or D lies in Q. Figure 5.12 PROOF: Since Q i P then it contains a straight line D' perpendicular to P (necessary condition for two planes to be perpendicular). D and D' being perpendicular to the same plane P are parallel. Consequently, D is either parallel to Q or lies in it because it is parallel to a straight line in this plane. I 76
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5.9 Theorem If a straight line D a plane P, every plane Q parallel to D P (use figure 5.12 for the proof). Given D P and Q parallel to D. Prove that Q P. PROOF: Since D is parallel to Q, D is parallel to a straight line D' in Q. Hence D P and D' is parallel to D which implies that D' P ( by the theorem that states if two straight lines are parallel, then every plane perpendicular to one of them is perpendicular to the other). D' P and D' is contained in Q Q is to P {sufficient condition for two planes to be perpendicular). Therefore, Q required. I 5.10 Theorem If two planes are parallel, then every plane perpendicular to one of them is perpendicular to the other plane also. Given Q is parallel to R and P Q. Prove that P R. Q Figure 5.13 77
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PROOF: Since Q P, then Q contains a straight line D P (necessary condition for two planes to be perpendicular). The plane Q being parallel to R implies that the straight line D in Q is parallel to R. The plane R being parallel to D which is perpendicular to plane P, is perpendicular toP (theorem 5.9). Therefore, P R. 5.11 Some Problems On Chapter Five 1. Consider a regular tetrahedron ABCD and let M be the midpoint of the edge CD. Show that the plane AMB the planes ACD and BCD. A D 8 c Figure 5.14 I SOLUTION: ABCD is a regular tetrahedron all the edges are equal. Since all the edges are equal all the faces are equilateral triangles. Since AACD is equilateral, the median AM CD. Similarly, BM CD. Hence CD AM and CD BM plane ABM. Since ABM, CDc plane ACD, and CD plane ABM plane ACD and plane ABM plane BCD. I 78
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2. Consider a plane P, a circle c lying in P, a diameter AB of circle C, a straight line D perpendicular to P at A and a straight line a cutting P at B. The straight lines D and a are not coplanar. Let M be any point on the circumference of c. Show that the planes (M,D) and (M,a) are perpendicular. Figure 5.15 PROOF: Since AB is a diameter of circle C LAMB=90 MB MA. D i MB c P. Then MB i plane (M,D). MB i plane (M,D) and MB c plane (M,d) plane (M,D) i plane (M,a) because plane (M,a) contains a straight line MB i plane (M,D). I 79
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6. Projections on A Plane 6.1 Projections Parallel To A Given Direction 6.1.1 Projection Of A Point Consider a plane P and a straight line A not parallel to P. The projection of a point A on the plane P along A is the point A' where the parallel to A, drawn through A, cuts the plane P. The plane P is called the plane of projection; AA' is called the projector or projecting line of point A. We notice that every point A f P has one and only one projection A' on P. The projection of a point B E P is the point B itself. We say that any point in P is unvariant in the projection. When the straight line A is perpendicular to P, the projection is called an orthogonal projection. Figure 6.1 6.1.2 Projection Of A Figure A Consider a figure (F). Associate to every point A E (F), its projection A' on a plane P along a given direction A. The set 80
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of all pointe A' thus obtained is a figure (F') which is the projection of (F) on P along A. 6.1.3 Theorem The projection of a straight line D on a plane P along a given direction A is, in general, a straight line 0'. Af Q A M D A I l N M' n Figure 6.2 PROOF: If D is parallel to A, its projection on P is reduced to a point. Suppose that D is not parallel to A. Through any point A on D draw AA' parallel to A. A' is the projection of A on P. The two intersecting straight lines D and AA' determine a plane Q parallel to A (if a straight line is parallel to a straight line in a plane Q then it is parallel to Q). The two planes P and Q intersect along a straight line D' passing through A'. The projecting line MM' of any point M on D lies completely in Q because if it does not, then this says that MM' cuts plane Q at M and, consequently, plane Q cuts A which is a contradiction because A is parallel to Q. Hence, the point is on D' because it belongs to the two intersecting plane P and Q. Conversely, the parallel to A drawn through any point M' on 81
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. I D' lies in Q and consequently, it cuts D at a point M whose projection isM'. Therefore, the projection of Don Pis the straight lineD'. (The plane Q is called the projecting plane of the straight line). 6.1.4 Special Cases I a) A straight line 0, parallel to a plane P, is parallel to its projection on P. b) The projection of a plane figure lying in a plane Q parallel to the given direction A lies on the line of intersection of the plane Q with the plane of projection P. 6.1.5 Theorem The projections of two parallel straight lines o1 and o2 on the same plane along a given direction A are, in general parallel straight lines. Figure 6.3 PROOF: If o 1 and o 2 are parallel to A, then their projections are reduced to two points. If o1 and o 2 lie in a plane parallel to A, then their projections coincide with the line of intersection of P and Q. 82
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Suppose that o1 and o2 are not parallel to A, and that they are not contained in a plane parallel to A. The projecting planes R and R' of these straight lines are parallel because two intersecting straight lines in one are parallel to two intersecting straight lines in the other (D1 parallel to o2 and AA' parallel to BB'). Therefore, the projections o1 and o2 of o1 and D2 are parallel since they are the intersections of two parallel planes by a third plane. I 6 .1. 6 Theorem The projection of a plane polygon on two parallel planes are congruent polygons. E A,f\.s I D I IE' A' _...,..._ D' "" B' I C' /, I I I _J...E: D" A" "cl. v " C" Figure 6.4 PROOF: Consider a plane polygon ABCDE whose along a given direction A, on two parallel planes P and P' are the polygons A'B'C'D'E' .... and A''B''C''D''E'' ... 83
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respectively. We have: A'B'B''A'' is a parallelogram because A'A'' is parallel to B'B'', and A'B' is parallel to A''B'' (A plane intersects two parallel planes along two parallel straight lines). Hence, the vectors A'B' and A''B'' are equal. Similarly, vectors B'C' and B''C'' are equal, as well as, C'D' and C' 'D' ..... Therefore, the angles of the polygons A'B'C'D'E' and A''B''C''D''E'' .... are respectively equal because their arms are respectively parallel and have the same sense. Also, their sides are respectively equal and, thus, these polygons are congruent. 6.2 Orthogonal Projection Of A Right Angle 6.2.1 Theorem I A necessary and sufficient condition for the orthogonal projection of a right angle, not having an arm perpendicular to the plane of projection P, to be a right angle is that it has at least one arm either parallel to P or lying in it. 6.2.2 Necessary COndition If the orthogonal projection of a right angle on a plane P is a right angle, then the given right angle has at least one arm either parallel to P or lying in P. Given the angle AOB=90'. Prove that AOB has at least one arm either parallel to P or lying in it. 84
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B Figure 6.5 PROOF: If OB is parallel to its projection O'B' or coincides with O'B', it is either parallel toP or lies in it. If OB cuts O'B', we have O'B' perpendicular to 00' because 00' is perpendicular to P and O'B' is perpendicular to O'A' by hypothesis, then O'B' is perpendicular to the plane (OAA'O'). Consequently, O'B' is orthogonal to OA which is contained in I the plane (OAA'O'). Now OA is orthogonal to O'B' and OA is perpendicular to OB, so OA is perpendicular to the plane (OBB'O'), and OA is perpendicular to 00' which is contained in the plane (OBB'O'). In the plane (OO'A'A), OA and O'A' are perpendicular to the same straight line 00', hence OA is either parallel to O'A' or coincides with it4 Therefore, OA is either parallel to P or lies in it. Thus, at least one of the arms of the angle AOB is either parallel to P or lies in it. I 85
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6.2.3 Sufficient Condition If a right angle has one arm either parallel to a plane P or lying in it while the other arm is not perpendicular to P, then its orthogonal projection on this plane is a right angle. Given angle AOB:90, OA is parallel toP and OBis not perpendicular to P. Prove that AA'O'B' is a right angle. B I Or I I I a' I I A' 0' Figure 6.6 PROOF: OA being parallel to P, it is parallel to its projection O'A' on this plane (one of the special cases of 6.1.4). Hence, O'A' is ortogonal to OB because OA is perpendicular to OB, and O'A' is perpendicular to 00' because 00' is perpendicular to P, so O'A' is perpendicular to the plane (O'B'BO). Consequently, O'A' is perpendicular to O'B' which is contained in the plane (O'B'BO), so AA'O'B' is a right angle. N.B.: We prove, in a similar way, that if OA lies in P, the orthogonal projection of AAOB on P is a right angle. 86
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6.2.4 Theorem If an angle has one of its arms parallel to a plane P or lying in it and its orthogonal projection on P is a right angle, then the angle itself is a right angle. Given AO is parallel to P and the angle A'O'B' is a right angle. Prove that the angle AOB is a right angle. B A./ I 01 j B' I ) I ) <'V A' 0' Figure 6.7 PROOF: Consider a right angle AOB having its arm OA parallel to P and whose orthogonal projection on P is a right angle A'O'B'. OA being parallel toP, it is parallel to its projection O'A' on this plane. But O'A' is perpendicular to the plane O'B'BO because it is perpendicular to two intersecting straight lines 00' and O'B' in this plane. Therefore, its parallel OA is perpendicular to this plane and consequently, OA is perpendicular to OB which is contained in OBB'O'. Hence the angle AOB is a right angle. I 87
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6.3 Angle Between A Straight Line And A Plane 6.3.1 Definition The angle that a straight line D makes with a plane P is the acute angle that this straight line makes with its orthogonal projection D' on this plane. If a designates the angle that the straight line D makes with the plane P, then tan a is called the slope of the straight line D with respect to the plane P. Figure 6.8 6.3.2 Theorem The acute angle that a straight line D makes with its orthogonal projection D' on a plane P is smaller than the angle that this straight line makes with any other straight line in P .. D' PROOF: Consider a plane P, a straight line D cutting P at Figure 6.9 0, any straight line 4 in P and the parallel Oz to 4 drawn through o. Let D' be the 88
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orthogonal projection of D on P, A' the orthogonal projection of any point A e D on P, and H the orthogonal projection of A on Oz. Let a= AAOA' and = AAOH. we want to show that In the right triangles AOA' and AOH, we have sin a = AA' AO and sin II AH AO (6.1) But AA'
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B I I I __.n_. a A' Figure 6.10 I 6.3.4 Theorem If two planes P and Q intersect, the straight lines in Q which are perpendicular to the intersection of P and Q make the greatest angle with P. Figure 6.11 PROOF: Consider two planes P and Q intersecting along xy, the perpendicular AB from a point A in Q to xy and any oblique AC to xy. Let A' be the orthogonal projection of A on P. Designate by a=AABA' and the angles that AB and AC make with P respectively. In the right triangles ABA' and ACA', we have 90
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sin a: AA' AB and sin = AA' AC but AB sin" > sin (6.5) AB AC and since the angles a and p are acute, we have a > p. N.B.: The straight lines of the plane Q which are perpendicular to xy, are called the lines of greatest slope of Q with respect to P. I 6.4 Orthogonal Projection Of A Polygonal Surface 6.4.1 Theorem The area of the orthogonal projection of a plane polygon on a plane is equal to the area of the polygon multiplied by the cosine of the acute angle formed between the plane of the polygon and the plane of projection. Designate by S the area of the polygon and by s the are of its orthogonal projection. 6.4.2 Area Of The Orthogonal Projection Of Triangle ABC i) The triangle ABC has one side BC lying in the plane of projection P. Let A'BC be the orthogonal projection of the triangle ABC on the plane P and AH the height of this triangle on BC. AA' is perpendicular to P AA' is orthogonal to BC which is in P. Now, BC is perpendicular to AH which is in the plane AA'H and BC is orthogonal to AA' which is in the plane AA'H, so BC is 91
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A B A' Figure 6.12 perpendicular to the plane AA'H which implies that BC is perpendicular to A'H. Hence the angle a is the plane angle of the dihedral formed by the plane P and the plane of the triangle ABC. Since s=area of A'BC=l/2(BCA'H) and S=area of ABC=l/2(BCAH), s s But in the triangle AA'H, we have cos a: = A1H s AH S COS IX S 92 (6.6) s cos tx (6.7)
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ii) The triangle ABC has one side BC parallel to the plane of projection P. Draw through BC the plane P' parallel to P. The projection of the triangle ABC on the planes P' and Pare congruent (theorem 6.1.3). Hence according to the preceding case, we have s :: S cos a (6.8) iii) The triangle ABC has no side lying in the plane of projection P or parallel to this plane. Draw through the nearest vertex B to P the plane P' parallel to P. Produce the side AC to cut p at a point D. The pointe A,C, and D being collinear, their projections A',C',and D on Pare also collinear {theorem 6.1.1). According to the two preceding cases, we have area A1BD = (area ABD) cos o: area BC1 D = (area BCD) cos o: Subtracting, we get But area A 1BD area BC1 D = (area ABD area BCD) cos a area A 1BD area BC1 D = area A 1BC1 = s area ABD area BCD = area ABC = S Therefore, s = s cos a. 93 (6.9) (6.10) (6.11)
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A p b Figure 6.13 6.4.3 Area Of The Orthogonal Projection Of Any Polygon Decompose the polygon into triangles. Designate by 51, 52, .. 5n the areas of the triangles thus obtained and by s 1 s 2 sn the areas of their orthogonal projections on P respectively. We have Sl = 51 COS cr. s2 = 52 cos cr. Adding, we get That is s = S cos a. 94 (6.12) (6.13) I
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Figure 6.14 6.5 The COmmon Perpendicular 6.5.1 Definition B I D' _,... .. c The common perpendicular to two straight lines in the space, is a straight line meeting each one of them at a right angle. 6.5.2 Theorem Between two skew straight lines, there exists one and only one common perpendicular, and the segment determined by D and D' on this common perpendicular is the shortest line segment joining a point of one of them to a point of the other. 6.5.3 Existence Consider two skew straight lines D and D'. Draw through D the plane P parallel to D', and through D' the plane Q perpendicular to the plane P thus constructed. The two planes P and Q intersect in a straight lineD''. 0' and D'' are coplanar in plane Q and at the same timeD' and D'' have no 95
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points in common because if they do then D' will not be parallel toP, which is a contradiction. Hence, D' and D'' are parallel. The straight lineD'' cannot be parallel to D, otherwise D' will be parallel to D which is impossible, because D and D' are skew. Therefore, D'' and D intersect at a point A. P and Q being perpendicular, the perpendicular at A to P lies in Q (theorem 5.5). This perpendicular cuts D' at B because it cuts its parallel D'' which lies in Q. We have AB perpendicular to P which implies that AB is perpendicular to D which lies in P and AB is perpendicular to D'' which lies in P. Hence, AB is perpendicular to D' because it is perpendicular to its parallel D'', and consequently, AB is a common perpendicular between the two straight lines D and D'. B Q Figure 6.15 M D 6.5.4 AB Is The Only Common Perpendicular Suppose MN is another common perpendicular between D and D'. MN being perpendicular to D', it is orthogonal to its parallel D''. Hence MN is perpendicular toP because it is orthogonal 96
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to two intersecting straight lines D and D'' of this plane. Draw ME perpendicular to 0''. Since P and Q are perpendicular, the straight line ME which lies in Q perpendicular to their line of intersection D'' is perpendicular toP. so, we have two perpendiculars MN and ME to P from the same point M which is impossible. Consequently, MN cannot be a common perpendicular to D and D', and AB is the only common perpendicular. 6.5.5 AB Is The Shortest segment Let E and F be any two points in D and D' respectively and FH the perpendicular drawn through F to P. The quadrilateral ABFH is a rectangle because its opposite sides are parallel and its angle at B is a right angle. Hence AB=HF. But FH
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! I 6.6 Some Problems On Chapter Six 1. Consider two parallel planes P and P', a fixed point o in P, and two points M and N, lying in P and P' respectively such that the angle OMN is a right angle. a) Find the set of N when M is fixed. b) Find the set of M when N is fixed. c) Find the set of M and that of N when the straight line MN varies but is always passing through a fixed point S. M d Figure 6.17 PROOF: a) b) OM is fixed because 0 and M are fixed. NM varies such that it is perpendicular to OM at the fixed point M the set of NM is a fixed plane Q perpendicular to OM at M. Therefore, the set of N is the line of intersection d of Q and P'. Let 0' be the orthogonal projection of N on P. Since N fixed, then 0' is fixed. NO' is perpendicular to p by construction, so NO' is orthogonal to OM which is in P. Now, OM orthogonal to NO' and OM perpendicular to MN (by 98 is
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hypothesis) implies that OM is perpendicular to the plane NO'M, OM is perpendicular to the plane NO'M OM is perpendicular to O'M which implies that the angle OMO' is a right angle. Therefore, the set of M is a circle diameter 00' lying in plane p, .;"fM ' 0 N Figure 6.18 c) Let SS' be perpendicular to P, then S'S produced intersects P' at S'', and since Pis parallel toP' and SS' is perpendicular toP, then SS'' is perpendicular toP'. Since Sis fixed, the two orthogonal projections S' and S'' on P and P' respectively are also fixed. SS' is perpendicular to P SS' is perpendicular to OM which lies in p, Now OM is orthogonal to SS' and OM is perpendicular to SM by hypothesis, so OM is perpendicular to the plane SS'M. OM is perpendicular to the plane SS'M is perpendicular to S'M that lies in the plane SS'M which implies that the angle OMS' is a right angle. Therefore, 99
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the set of M is a circle diameter OS' lying in plane P. Since SS'' is perpendicular toP', then SS'' is perpendicular toNS'' that lies in P' which implies that the set of N is NS'' because it is a fixed straight line that is perpendicular to fixed S'S' at a fixed pointS''. Figure 6.19 I 2. A triangle ABC, right angled at B, lies in a plane P. A point S is taken on the perpendicular to plane P at A. What is in the figure and why: a) The angle that SB makes with plane P? b) The line of greatest slope of the plane SBC with respect to the plane? c) The common perpendicular between SA and BC? 100
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I J SOLUTION: a) The projection of SB on plane P is AB, and the angle that SB makes with plane P is the acute angle that SB makes with its orthogonal projection on P. Such an angle is SBA. s c B Figure 6.20 b) SA is perpendicular to P SA is orthogonal to BC which lies in P. Since BC is orthogonal to SA and BC is perpendicular to AB, then BC is perpendicular to the plane SAB which implies that BC is perpendicular to SB. Now planes SBC and P intersect along BC, and SB is in plane SBC and perpendicular to BC. So SB is the line of greatest slope of the plane SBC with respect to plane P. c) SA is perpendicular to P SA is perpendicular to AB which lies in P. So, AB is perpendicular to SA and AB is perpendicular to BC (by hypothesis). Therefore, AB is the common perpendicular between SA and BC. I 101
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3. consider two skew straight lines D and D', their common perpendicular 00'=4a (0 is on D and 0' is on D'), a point M on D such that OM=x and a point M' on D' such that O'M'=2x. calculate MM' in terms of a and x in the following cases: a) D and D' are orthogonal b) D and D' make and angle of 45 with each other. 0 X M 0 D' Figure 6.21 SOLUTION: Through 0' draw D'' parallel to D. Drawn MM'' parallel to 00' to cut D'' at M''. Then Dis perpendicular to 00' and D'' is parallel to D, soD'' is perpendicular to 00'. Since 00' is perpendicular to D'' and 00' is perpendicular to D', 00' is perpendicular to the plane MM'' is parallel to 00' MM'' is perpendicular to the plane (D',D'') which implies that MM'' is perpendicular to M''M'. Therefore, the triangle MM''M' is rightangled at M''. 102
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a) Now, since D' and Dare orthogonal, '0'0'' is a right angle, which means in the right triangle O'M'M''. In the right triangle MM''M', we have MM72 = MM112 + M"M" = 16 a' sx' MM1 = ../16a2 + 5x2 sx' (6.14) (6.15) b) In this case, the angle D'O'D'' is 45 and we can apply the law of cosines on the triangle M'O'M'': M 1 M112 = x' + 4x2 2 (x) (2x) cos 45" = Sx' 4x2 12 = 5x2 s,f2 x' 2 (6.16) Therefore, in the right triangle MM''M, we can write MM 1 = MM17' + MT;M" = 16a2 + sx' 2/2 x' MM 1 = Y16a2 + 5x2 2/2 x' 103 (6.17) I
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7. General Problems 1. a) Given a gauche quadrilateral ABCD. Let M,N,P and Q be the midpoints of AB, BC, CD, and DA respectively. Show that MNPQ is a parallelogram b) Deduce that the line segments which join the midpoints of the opposite edges of a tetrahedron are concurrent at a point 0 which is the midpoint of each one of these line segments. A B D c Figure 7.1 PROOF: a) In the triangle ABC, we have {7 .1) since M and N are the midpoints of AB and BC. Similarly, in the ACD, we have 104
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Therefore, = MNPQ is a parm. b) Since MNPQ is a parm the diagonals MP and NQ bisect each other at 0. Similarly, we can prove that LQJN is a parm. (the proof is similar to 7.l.a) [Here we have L and J as midpoints of AC and BD respectively]. Since LQJN is a parm. LJ and NQ bisect each other at 0. Therefore, MP, NQ and LJ are concurrent at 0 which is the midpoint of each of these segments. I 2. Consider a tetrahedron ABCD in which the edges AB and CD are orthogonal, AB=CD=a and BC=b. A plane parallel to AB and CD cute the edges Be, AC, BD and AD in M,N,P and Q respectively. a) Show that MNQP is a rectangle. b) Calculate the area of MNQP in terms of a,b and BM=x. c) For what value of x would this area be maximum? Figure 7.2 105
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PROOF: a) AB and NM are coplanar in plane ABC. Also, AB and NM have no points in common because AB is parallel to the plane MNQP which contains NM. Therefore, AB and NM are parallel. Similarly QP is parallel to AB, therefore, NM is parallel to QP (if two straight lines are parallel to a third, then they are parallel to each other). In a similar fashion, we can prove that NQ is parallel to MP, therefore, MNQP is a parm. Since AB and CD are orthogonal ANMP is a right angle because ANMP is the angle between AB and CD. Therefore, MNQP is a rectangle. b) The triangle CMN is similar to the triangle CAB and we can write CM = CB MN bbx BA MN MN = a(bx) a b ABMP is similar to ABCD and we can write EM = MP_x=MP_MP=ax CD b a b Therefore, 106 (7. 3) (7 .4)
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Area MNQP = MN MP = a(bx) b 2 = E._ x(b x) units' b' ax b (7.5) c) Area is max .... (bxx2 ) b' is max. Since a' is constant b' then (bxx2 ) is max when (bx x2 ) is max 2x + b = b' 0 x = Geometrically, this says that the area is max when M is the midpoint of BC. 3. A parallelogram ABCD lies in a plane P. Let Ax, By, cz and Dt be parallel semistraight lines drawn from A,B,C and D to the same side of P. A plane P' cuts Ax, By, Cz and Dt at A', B', C' and D' respectively. a) Show that A'B'C'D' is a parallelogram. b) Show that AA' + CC' = BB' + DD'. c) State and prove the converse of part b. t z y c :::::. C' c Figure 7.3 107
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PROOF: a) Ax and Dt determine a plane (Ax,Dt) because they are parallel. Similarly By and Cz determine a plane (By,Cz). Plane P' intersects planes (Ax,Dt) and (By,Cz) along A'D' and B'C'. Plane (Ax,Dt) is parallel to plane (By,Cz) because two intersecting straight lines Ax and AD in plane (Ax,Dt) are parallel to two intersecting straight lines By and BC in plane By,Cz. Therefore, A'D' is parallel to B'C' (if two parallel planes are cut by a third plane, the lines of intersection are parallel). Similarly, A'B' is parallel to D'C'. Therefore, A'B'C'D' is a parm because the opposite are parallel. b) Join the diagonals of the two parma. The diagonals A'C' and B'D' bisect each other at 0'. Also, the diagonals of ABCD bisect each other at 0. AA1+cc' From plane geometry, 001 = in the quad AA'C'C. therefore, AA1+CC1 2 2 BB1 +DD1 Also,oo' = in the quad DBB'D', 2 BB'+DD'AA1+CC1 = BB1+DD1 2 (7. 6) c) The converse is: if we choose points A',B',C' and D' on Ax, By, Cz and Dt respectively such that AA' + CC' = BB' + DD', 108
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then A',B',C' and D' will be coplanar and A'B'C'D' is a parallelogram. PROOF: A',B' and D' are three noncollinear points, and hence they determine a plane (A'B'D') which cuts Cz at C''. Then we have a plane A'B'C''D'. According to part a we can say that A'B'C'1D1 is a parm, and according to part b, we can say that AA I+ cc" = BB1 + DD1 But AA, + cc I = BB, + DO I therefore I AA' + CC'' = AA1 + CC' CC'' = And, since and C'' are collinear, C' and coincide. So, A',B', C' and D' are coplanar and from part a, the quad A'B'C'D' is a parallelogram. I 4. Consider a plane Q and a point Sf Q. We call perspective of a point M the point M' where the straight line SM cuts the plane Q. Consider a fixed plane P which cuts Q along a straight line D. What are the points of P which do not have perspectives? D' /. s. L!:L Figure 7.4 109
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PROOF: Through S draw a plane P' parallel to Q. P' cuts P along a straight lineD'. D' and Dare coplanar in plane P. Also, D' and D have no common points because if D meets D', then D will be meeting P' where D' lies, which says that P' and Q are not parallela contradiction. Hence, D' and D are coplanar and do not have any points in common D is parallel to D'. Now, any point Me D' has no perspective because SMc P' SM I Q SM does not cut Q. Hence, the points on D'c P have no perspectives. I 5. An equilateral triangle ABC lies in a plane P, M is a point on the perpendicular to P at the center 0 of the triangle ABC. Show that the opposite edges of the tetrahedron MABC are orthogonal Figure 7.5 PROOF: 0 is the center of the equilateral triangle ABC 0 is the point where the heights of the triangle ABC concur, so 0 is the orthocenter of triangle ABC. Now, OMJ.POMJ.BCcP, so 110
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OM and AH imply that BC is perpendicular to the plane AMO because BC is orthogonal to two intersecting straight lines OM and AH in this plane. Then BC AMOBC MA the two opposite edges MA and BC of the tetrahedron MABC are orthogonal. Similarly, we can prove that AB MC, and MB AC. Hence, the opposite edges of MABC are orthogonal. I N.B.: See problem 6 which is a continuance of problem 5. 6. In problem 5, if the perpendicular at M to the plane MBC cuts Pat a point A', show that the three points A, A' and 0 are collinear. Also prove that oFf = OA 1 OH. M 0 Figure 7.6 PROOF: Using the figure in problem 5, plane (MBC) MA' BC. MA BC (We proved this in problem 5). MO PMO BC. Hence MA', MA and MO are straight lines drawn through M orthogonal to BCMA', MA and MO lie in a planeR which is drawn through M perpendicular to BC. Hence A', A and o are 111
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common to planes P and R A', A and 0 are collinear (points common to two intersecting planes belong to the line of intersection of the two planes). As to the proof of the second part, MA' plane (MBC) MA' MH c plane (MBC). MO P MO A'H. The problem now is a plane geometry problem, and specifically a relation in a right triangle A'MH where MO is the height on A'H, and this relation is OFf = OA 1 OH I 7. Consider two circles c and C', of radii Rand R', of centers 0 and 0', and having the same axis 00'. Let A be a point on the circumference of C and A' a point on that of C'. a) Construct a pointS on 00' equidistant from 0 and 0'. b) Show that s is equidistant from all the points of C and C'. I A A '< I S !.1 I Q 8?, :A' (C') Figure 7.7 112
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PROOF: a) Draw the mediator plane Q of AA'. Q cuts 00' at s. Since SeQ* SA= SA', then Sis the required point. (7.7) b) Sis on 00', the axis of the circle, *Sis equidistant from all the points of c. Similarly, Sis on 00', the axis of C', Sis equidistant from all the points of C'. Now, since SA=SA', then Sis equidistant from all the points of C and C'. I 8. Consider a tetrahedron OABC in which angle AOB is equal to angle BOC is equal to the right angle COA. Let H be the foot of the perpendicular drawn from 0 to the plane ABC. a) Show that H is the orthocenter of the triangle ABC. b) Show that A L' Figure 7.8 1 oJF + 0 B 113 1 oF + 1 x c (7 .8)
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PROOF! a) By hypothesis, OA .L OB OA .L plane (OBC) (7. 9) OA .L OC OA .L plane (OBC) AO .L BC c plane (OBC) (7 .10) OH .L plane (ABC) OH .L BC c plane (ABC) (7.11) Now, BC .L OH BC .L plane (OAH) (7 .12) BC .L OA BC .L plane (OAH) BC .L AL c plane (OAH) (7.13) Therefore, AL is a height in the triangle ABC. Similarly CL' is a height in ABC. so, H is the orthocenter of the triangle ABC because it is the point of intersection of two heights in ABC. b) OA .L to the plane OBC (proved above) OA .L OL c plane OBC AAOL is rightangled at L. Since OH .L plane ABC OH .L AL c (ABC). Therefore, the triangle AOL is a right triangle and AH is the height relative to the hypotenuse AL. Hence, we can write the following relation: 1 OJf' 1 OA2 + 1 01] (7 .14) (relation in a right triangle). BC is perpendicular to plane OAH (proved above) BC is perpendicular to OL which is in the plane OAH. Therefore, triangle BOC is right114
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angled at 0 and OL is a height relative to the hypotenuse BC. Now, we can write 1 OL2 = 1 + of32 1 OC"(7 .15) (relation in a right triangle). Substitute (7.15) into (7.14) to get 1 o& = 1 OA2 + 1 o& + 1 OC"(7 .16) I 9. Given a straight line D and a point 0 D. Let Q be a variable plane passing through D and denote by H the foot of the perpendicular drawn from 0 to Q. a) Find the set of the straight lines OH b) Find, under the same conditions, the locus of H. D Figure 7.9 115
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PROOF: a) OH .1 Q OH .1 D c Q. o is fixed, D is fixed and OH is a variable straight line orthogonal to D the set of OH is a plane Q' drawn through 0 perpendicular to D. b) H is common tc two fixed planes Q and Q' the locus of H is the line of intersection of Q and Q'' i.e., d = Q n Q'. I 10. (Hard Problem) Consider a fixed circle 0 in a plane P. A is a fixed point external to P. BC is a variable chord of the circle (0) so that = 90. Find the locus of M, the midpoint of BC. A II Figure 7.10 116
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SOLUTION: Draw AH i P. A is fixed H is fixed AH = constant = d. OH = constant = 2a. Since M is the midpoint of BC OM i BC AOMC is rightangled at M (7.17) where R is the radius of circle (0). AABC is rightangled at A the median AM relative to the hypotenuse BC is (BC)/2 CM = MA. Therefore, Mo' = R2 Me" = R2 MA2 AH i P AH i HM AAHM is rightangled at H MA 2 =Aif' + mf = d2 + HM' Mi52 = R2 MA2 = R2 (d2 + HJ.f) = R2 d2 HM' M3' + MJi2 = R' d2 = constant Let I be the midpoint of OH Mo' + MJ{' = 2IM' + OJi' (relation in a A) 2 = 2IM' + 4a2 = 2IM' + 2a2 2 R2 d2 = 2Iif + 2a2 2IM' = R2 d2 2a2 I R2 d2 2 2 IM = 2 a = constant 117 (7.18) (7 .19) (7 .20)
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Since OH is fixed I, the midpoint of OH, is fixed. Since I is fixed and IM = constant, then the locus of M is a circle whose center is I and whose radius is IM = (7.21) and this circle lies in the plane P. I 11. (Long problem) Given two axes x'x and y'y orthogonal to each other. Let AA' be their common perpendicular (A in x'x and A' in y'y) and 0 the midpoint of AA'. Let AA' = 2a, P a point in x'x and P' a point in y'y. M is the projection of 0 on PP' and H is the projection of 0 on A'P. a) Suppose, in this part, that P is fixed and P' traces y'y. i) What is the direction of the straight line A'P' with respect to plane (AA'P)? Prove that the two planes (AA'P) and (A'P'P) are perpendicular. Show that OH is perpendicular to plane (A'P'P). ii) Find LHMP. Then deduce the set of M as P' moves on y'y. iii) Calculate, in terms of a, the power of the point A' with respect to the circle lying in the plane (A'PP') and whose diameter is HP. Show that this power is independent of the position of P on x'x. 118
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b) Suppose now that P and P' vary simultaneously such that AP = A 1 P 1 constantly. i) Compare OP and OP' ii) What is the position of M on PP'? iii) Find the set of M. c) Suppose now that P and P' vary simultaneously so that AP = x and A 1 P 1 = 2a x. Between what limits should x vary? y Figure 7.11 SOLUTION: a) i) A'P' 1. AA'i I A'P' 1. plane (AA'P) A'P' 1. AP J (7 .22) plane (A'P'P) 1. plane (AA'P) because plane (A'P'P) contains the straight line A'P' which is perpendicular to plane (AA'P). OH 1. (A'P'P) because if two planes are then every straight line in one of them to the line of intersection of the two planes is 1. to the other plane. 119
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ii) OH i (A'P'P) OH i MP (7 .23) MP i OH MP i plane (OHM) MP i HM c plane (OHM) MP i OM (7 .24) and, therefore, AHMP = 90. Hand Pare fixed and AHMP = 90 the set of M is a circle diameter HP lying in the fixed plane determined by the fixed point P and the fixed straight line y'y. iii) The power of A' with respect to circle (HMP) =@ A'/(HMP) = A'H x A'P (definition of power of a point with respect to a circle). (AOHP) is a cyclic quadrilateral because LOAP + LOHP = goo + 900 = 180o A1/(0APH) =the power of A' with respect to circle (OAPH) =A 1HxA 1P=A 10xA 1 A = ax2a = 2a 2 =constant (7 .25) Therefore,@ A'/(HMP) = 2a2 @ A'/(HMP) is independent of the position of P on x'x. b) i) Consider and AP = A'P' OA = OA' AOAP : OP = OP' t..OAP=t..OA'P'=9Q (7.26) ii) OP = OP' is isosceles the height OM bisects P'P M is the midpoint of PP'. 120
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__________ 7 P __ x Figure 7.12 iii) Through A' draw A'Yix'x. Draw PP"IAA'. AA' L A'Y I f AA' L plane (XA'Y) AA' L A'X ) AA' L plane (XA'Y) I X } PP" L plane ( XA' Y) PP" AA' J (7. 27) (7 .28) MM' I PP" MM' L plane (XA'Y) M' is the orthogonal projection of M on plane (XA'Y). Let the coordinates of M' be X and Y with respect to the system of axes (X'A'X, Y'A'Y).(Figure 7.13) (7.29) (7 .30) But A 1 P 1 = A 1 P 0 X= Ythe set of M' is the bisector of the angle LXA'Y, whose equation is Y = X. But M' is the orthogonal projection of M on plane (XA'Y) 121
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the set of M is a straight line lying in the mediator plane (ZOZ') of AA', and this straight line is the bisector of A.ZOZ' [OZ I x'x, OZ' I y'y]. (7.31) AP = X, A I pi = 2a X (7. 32) A 1 P 1 2 0 2a X 2 0 2a 2 X (7. 33) Combining (7.32) and (7.33), we get: 0 x 2a (7 .34) y P" F M'(X,Y) X' X A' P" y Figure 7.13 12. Consider two orthogonal straight lines D and D', and their common perpendicular AB =a [A e D, BeD']. A point M e D such that AM = y, and a point N e D' such that BN = x. I is the midpoint of MN. Through B draw BY 122
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parallel to D. Denote by I' the projection of I on the plane (D',BY) determined by D' and BY. Find the Set of I' in the following cases: i) M and N vary such that AM BN= k (k being a constant) ii) M and N vary such that AM+ EN= 1 (1 being a given length) iii) M and N vary such that AM' BN'= d2 (d is constant). A y M D X Figure 7.14 SOLUTION: Draw MM' I AB. Then M'N is the orthogonal projection of MN on plane (D', BY). I' is the midpoint of M'N. BM1 =y, BN =x. Consider the orthonormal system (BY ,X' BX), where X'BX is D' itself. Let the coordinates of I' be X andY. 123
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I I .I i) y Figure 7.15 X= BN X = X= 2X 2 2 Y= BM1 1' y = 21 2 2 AMBN=kyx=k 2 Y 2X = k Y = K 4X *the set of I' is a hyperbola whole equation is k Y = lying in the plane (X' BX, Y 'BY). 4X ii) Using the above axes: y+x=1 2Y + 2X = 1 2Y = 2X + 1 Y = X+ 1 2 124 (1.35) (7.36) (7 .37) (7. 38) (7.39) (7 .40) (7. 41)
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the set of I' is a straight line whose equation is Y =X+ l/2 lying in the plane (X'BX,Y'BY). d' 4 the set of I' is a curve whose equation is d' 4 (a hyperbola). (7. 42) (7. 43) I 13. In a plane P, consider two perpendicular semiaxes ox and oy. Through o, draw Oz perpendicular to P. Consider a point S in the plane (yOz) whose orthogonal projection on Oz is A and on Oy is B such that Wi = a and OB = 2a, where a is a given positive constant. Let M be a variable point on ox and H the foot of the perpendicular drawn from 0 to the straight line AM. Take OM = x. a) Find the locus of H. b) The straight line SH cuts the plane (xOy) in a point N. Find the locus of N. 125
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SOLUTION: X Figure 7.16 z a) OA is fixed and LOHA = 90 the locus of H is a circle T, diameter OA, lying in the plane (xOz). b) SA I Oy SA (xOy) the plane (SANM) cuts plane (xOy) along MN I SA. The two triangles MNH and SHA are similar 2a + MN HA + MH 2a + MN MO is tangent to the circle T x' is right angled at 0 Now, Md'=MHMA =MfiMAMH= x' MA 126 MH MA (7. 44) (7 .45) {7 .46)
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MN 2a + MN MH MA ;;::'MN':";;; 2a + MN MN = 2ax2 = 2x2 WI = a' a 2x' a x' MA' x' (7.47) Oy (xOz) MN which is parallel to Oy is to plane (xOz) MN ...1.. Ox ... MN = y = the ordinate of N and OM = x ::. the abscissa of N. Therefore, MN = 2 X2 y = 2 X2 the set of N is a parabola a a whose equation is y = x', and this parabola lies in plane a (xOy). I 127
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Bibliography AltshillerCourt, Nathan. Modern Pure Solid Geometry. New York: Chelsea, 1964. M.N., and William Wernick. Problems and Solutions in Euclidean Geometry. New York: Dover, 1968. Hart, Walter W., and Veryl Schult. Solid Geometry. Boston: Heath, 1952. Leighton, Henry L.C. Solid Geometry. New York: Nostrand, 1947. Lines, L. Solid Geometry. London: Macmillan, 1935. Richardson, Sophia Foster. Solid Geometry. Boston: Ginn, 1914. Seymour, F. Eugene, and Paul James Smith. Solid Geometry. New York: Macmillan, 1949. Sigley, Daniel T. and William T. Stratton. Solid Geometry. New York: Dryden, 1956. Simpson, J.L. Solid Geometry. New York: Harper, 1960. 128

