BOOK XT. PROP, xvii. xxi.] MATHEMATICS. PLANE GEOMETRY. 



695 



PROPOSITION XVII. THEOKEM. 

 If two straight lines (AB, CD) are cut by parallel planes 

 (G H, K L, M N, tn points A C, E F, B D), they are 

 cut in the same ratio, (i. e., A E : E B : : OF: 

 FD). 



For, join A D meeting K L 

 in X, and join A C, E X, X F, 

 and BD. Then, because the 

 plane ABD cuts the parallel 

 planes K L, MN in the lines 

 E X and B D, these are paral- 

 16X1. lels.* Similarly AC 

 is parallel to XF ; now because 

 E X is parallel to B D, and 

 A C to X F, we have, 

 AE : EB :: AX : XDf 



HTL 



andAX: XD : : CI+FD 

 .'. AE: FB :: CF : FD. Q. E. D. 



PROPOSITION XVIII. THEOREM. 

 If a straight line (A B) 6 at right anglts to a plane (C K), 



every plane that passes through the line is at right 



angles to the plane. 



Let C H be any plane passing through A B, and let 

 it intersect C K 

 in the line C E. 

 From any point 

 F in C E, and 

 in the plane 

 C II, draw F G 

 at right angles / 

 .111. toCJJ.* / 

 Then, becavse f 

 ABU perpen- /--'-^ 

 dicuiar to the 



iDcf.xi. plane it is perpendicular to CE,t .". ABF 

 is a right angle. But G B F is also a right angle, /. 

 G F B and F B A are together equal to two right angles ; 



MI. ' GF U parallel to B A,* and ABU per- 

 pendicular to the plane C K, .'. G F is perpendicular to 



. H xt. the plane C K;t similarly any other line in 

 C H drawn perpendicular to C E is perpendicular to the 

 plane C K, .'. plane C H U perpendicular to plane C K. 

 In like manner it can be proved that any other plane 

 passing through ABU perpendicular to C K. Q. E. D. 



PROPOSITION XIX. THEOREM. 



If two planes (A B, B C) which cut one another (in the 

 line B D) be each perpendicular to a third plane (A D C), 

 the common section (B D) is also perpendicular to the 

 same plane. 

 Let plane A B intersect plane A D C in A D, and let 



plane B C intersect plane 



ADC in DC. Then, if , 



B D U not perpendicular to 



A D C, from point D in 



plane A B, draw D E per- 



11 i. pendicular to AD,* 

 and similarly in B C draw 

 D F perpendicular to D C. 

 Then, because plane ABU 

 r>eri*mdieular to A D C, we 



nave ED perpendicular to "= 



+ Def. XI. A D C,t similarly FD is perpendicular to 

 A DC; .'. from the point A D in the plane A DC, two 

 lines D E, D F are drawn perpendicular to that plane, 



is xi. which U absurd :* .'. B D U perpendicular to 

 the third plane A D C. Q. E. D. 



PROPOSITION XX. THEOREM. 

 If a solid angle (at A) be contained by three plane angles 



(B A C, C A 1), DA B), any two of them are greater 



tluin the third. 



If the three plane angles are equal, any one of them 

 U clearly less thu the other two. 



\ 



\ 



If the three angles are not equal, let B A C be that 

 which is not less than either of the other two. 



At A in BA, and in 

 plane B A C, make the 

 angle B A E = angle 



231. B A D,* make 

 AE = AD, through 

 E draw B E C, meeting 

 A B and A C in B and 

 C. Join D B, D C. 



Then in triangles B A D, B A E, the sides B A, A D = 

 the sides B A, A E each to each, and the angle B A D = 



+ 41. the angle B A E, .'. the base B D = base B E.f 

 Now, the sides B D, D C are together greater than B C ; 



20 i. t. e., greater than BE, C E ; taking away the 

 equals B D, B E, we have left D C greater than E C.t 



+ 5 AT i. Then in the triangles E A C, D A C, we have 

 the sides D A, A C = the sides E A, AC, each to each. 

 But the base DC is greater than the base EC, .'. the 



25 I. angle D A C U greater than the angle E A C.* 

 To each of these add the equals DAB, B A E. Then 

 the two BAD, D A C are together greater than B A C.* 



t 4 AX. I. Q. E. D. 



PROPOSITION XXI. THEOREM. 



Every solid angle (A) is contained by plane nn/les, which, 

 together, are less than four right angles. 



First, let the angle (A) be contained by three plane 

 angles B A C, CAD, DAB. Join B C, CD, D B. 

 Now, since the angles of 

 any triangle are together 

 equal to two right 



32 1. angles,* we have 

 the angles of any three 

 triangles together equal 

 to six right angles, .'. 

 BAC+CAD+DAB 

 + ABC + BCA + 

 ACD+CDA+ADB+ 

 D B A = six right angles. 



Now, because D U a solid angle, C D A + A D B are 

 + JOXI. greater than CD B.* Similarly D B A + 

 A B C are greater than D U C, and B C A + A C D are 

 greater than BCD. 



.-.BAC + CAD + DAB + CDB + DBC+BCD 

 are less than six right angles. Now, B C D is a triangle, 

 and .'. CDB+DBC+BCD=2 right angles, .'. 

 BAC + CAD + DAB are together less than four right 

 angles. But these contain the solid angle A, .'. when 

 three angles contain a solid angle, they are together less 

 than four right angles. Q. E. D. 



Next, suppose the angle A to be contained by any 

 number of plane 

 angles. Then all 

 these plane angles 

 are together less 

 than four right an- 

 gles ; for supposing 

 the lines containing 

 the angle to be cut 

 by a plane, so that 

 we obtain a polygon 

 B C D E F instead of the triangle in the first case. As 

 before, we can easily prove that the angles containing 

 the solid angle, together with the angles of the polygon, 

 are less than twice as many right angles as there are 

 sides of the polygon ; but twice as many right angles as 

 there are sides of the polygon are equal to the auglest 



Cor. 33 1. of the polygon, together with four right 

 angles ; .'. the plane angles forming solid angles at A, 

 together with angles of polygon, are less than angles of 

 polygon, together with four right angles, .'. angles form- 

 ing solid angles are together less than four right angles. 

 Q. E. D. 



The study of the last two piopositions will be of value 

 in connection with the section on Crystallography. ED. 



