I 



MATHEMATICS. PLANE GEOMETRY. [BOOK XL PROP, x. XVL. 



xi. 



or A B and C D are parallel to each other. 



PROPOSITION X. THIOKKK. 

 If tieo straight lintt (A B, It C) meeting one another be 

 parallel to tico olhen (D E, E F) alto mntin-i one another, 

 tht latter two not beiny in the tame plane with the former 

 two, then the farmer liru contain an anyle (A B C) equa 

 to the angle (D E F) contained by the latter two. 

 We suppose that ABU parallel to 1) K, and B C to 

 TakeBA-ED.andBC-EF, and join BE, 

 A D, C F, A C, D F. 



Then because ABU equal and 

 parall.4 to 1) K, and their extremities 

 are joined towards the same point 

 B E and A D ; .'. B E U equal am 



MI. parallel to A I).* Similarli 

 B E U equal and parallel to C F, ant 



+ AX u .'. AU U equalf and paral 



9 xi. lei* to C F. But the ex 

 tremities of A D, C F are joinet 

 towards the same points by A C, ani 



t w i. 1 ) F, and therefore A C - 1 ) F. t 

 Hence in triangles A B C, D E F we 

 hare the sides A B, B C = the side U E, E F, each to 

 each, and the base A C the base D F, .'. the angle 

 .l. ABC = the angle DBF.* Q. E. D. 



PROPOSITION XI. -PROBLEM. 

 From a giren point (A) abore a plane (B C H) to draw a 



hne perpendicular to that plane. 

 In the plane draw any line B C. From A draw A D 



11 1. perpendicular to B C. * In the given plane from 

 + ni. I) draw E 1) perpendicular toBC.t From A 



u i. draw A F perpendicular to ED,* then A F is 



the line required. 



From through F draw G H, 

 t si I. parallel to BC.f Now, 

 since B C is perpendicular to 

 E I) and A D, it is perpen 

 dicular to the plane passing 



4X1. through them,* 

 G H is also perpendicular to 

 the plane passing through E D 

 t 8 xi. and A D,t and .'. is perpendicular to FA;*.'. 

 i*f. xi. the angle A F G is a right angle. But, by con- 

 struction A F D is a right angle, then since A F is per- 

 pendicular to the lines FG, F D it is also perpendicular 

 1 4 xi. to the plane passing through them,t i. e. t is per- 

 pendicular to the plane B C H. Which was to be done. 



PROPOSITION XII. PROBLEM. 

 From a given point (A) in a given plane (E F) to draw a 



line perpendicular to that plane. 



Take any point B above the plane, and from B draw 

 BO perpendicular to the 

 ll XI. plane.* From 

 A draw A D parallel to 

 | t as I. B C ;t then be- 

 cause A D and B C are pa- 

 . c N. rallel, and B C is perpendi- 



cular to the plane, .'.AD 

 r. F is also perpendicular to the 



8 xi. plane.* Which was to be done. 



PROPOSITION XIII. THEOREM. 



From the. tame point (A) of a given plane, there cannot be 

 two ttraight lines (AB, AC) at right angles to the plane, 

 upon the tame, side of it ; and there can be but one per- 

 pendicular drawn to a plane from a given point above it. 

 For, if possible, suppose A B, AC to be at right angles 



to a given plane. Suppose the plane which contains AB 



and A to intersect the given plane in D E. Then A, 



B A, D A, are in the same 



plane. Now C A U perpen- 



dicular to every lino in the 



Df. xi. plane,* and .'. to 

 AD, .'. C A D U a right an- 



le. For the same reason 

 AD U a right angle, /. 

 A!.II i. CAD-BAD,t ' 



} 



gl 

 B 



which U absurd. Also from the same point above a 



> pi. mo, two perpendiculars cannot be drawn to it, 



for if they could they would be parallel to one another, * 



XL which U absurd. Q. E. D. 



PROPOSITION XIV. THEOEKM. 



Planet (D C, E F) to which the time ttraight lint (A B) it 

 perpendicular are parallel to each other. 



For if not, the planes must 

 intersect ; let them intersect in 

 the straight line H G, in which 

 take any point K. Join K A, 

 K I :. Then K A U in the plane 

 D C, .'. K AB u a right angle.* 



Df. XI. Similarly A B K U 

 a right angle, .'. the two angles 

 K A B, ABK of the triangle 

 A It K, are equal to two right 

 angles, which is absurd, t .'. 



i 17 i. the two planes cannot 

 intersect, and therefore are pa- 



Def. xi. ralleL* Q. E. D. 



PROPOSITION XV. THEOREM. 



If two straight lines (AB, BCJ meeting one another be 

 parallel to two other straight lines (D E, E F) whi<-h 

 meet one another but are not in the same plane with the- 

 former two, then the plane passing through the fornter 

 two (A B, B C) is parallel to the plane passing through 

 the latter two(DE, E F). 



For, from B draw B G perpendicular to the plane 

 ll xi. ABC,* meeting the plane DEFinG. In 

 this plane and through G, draw GH, GK, parallel to 

 1 31 1. D E, E F.f Then, since E D is parallel both 

 to A B and G H, .'. A B is 

 9x1. parallel to GH.* 

 Similarly G K is parallel to 

 B C. Again, because B G is 

 perpendicular to the plane 

 ABC, it is perpendicular to 

 + ppf. xi. AIt,f .'. ABG is 

 a right angle ; but because 

 A B is parallel to G H, A B G 

 and B G H are together equal 

 to two right angles, * .'. BGH 



29 1. is a right angle, Similarly B G K is a right 

 angle, .'. BG is at right angles to GH and G K, and 

 therefore is perpendicular to the plane passing through 



1 4 XL them ;f i.e., is perpendicular to the plane DEF. 

 Then, since H G is perpendicular to both the planes 



u xi. A B C, D E F, these planes are parallo'. * 

 Q. E. D. 



PROPOSITION XVI. THEOREM. 



If two parallel planes (A B, C D) are cut by a third plane 

 (EH), their intersections with it (E F, G H) are 

 parallels. 



For if not, they will meet when produced either on the 

 ide F H or E G ; let them be produced on the side FH, 

 and meet in K. Then, since G H U in the plane C D, 



G H when produced in in 

 l xi. that plane,* .'. 

 G H K U in the plane C D. 

 Similarly E F K is in the 

 plane AB, .'. the plane 

 A B meets the plane C D, 

 for they have a common 

 point K, and therefore is 

 not parallel to it, which 

 is contrary to the hypo- 

 thesis, .'. EFandG H do 

 not meet when produced 

 on the side FH. Similarly 

 hey do not meet when produced on the side E G. But 

 mes which are in the same planes, and being produced 

 ither w.iy do not meet, are parallel, .'. E F and G H are 

 arallel. Q. E. D. 



