BOOK i. PROP. XL xvi.] MATHEMATICS. PLANE GEOMETRY. 



545 







T. 9. C D ;t A B shall be cut into two equal parts 

 in the point D. 



Because C A = C B, and C D common 

 to the two triangles A C 1 ), BCD, the 

 two sides C A, C D are = the two C B, 

 C D, each to each ; and the angle A C U 

 = the angle BCD;.', the base A D = 

 Pr. 4 the base B D;* . . the given line 

 A B is dividfd into two equal parts in tlie 

 point D. Which was to be done. 



PROPOSITION XL PROBLEM. 

 To draw a straight line at right angles to a given straight 



line (A B), from a giren point (C) in it. 

 Take any point D in A C, and make C E= C D. Upon 

 _ D E describe an equila- 



teral triangle DFE,* 

 Pr. i. and draw C F ; 

 CF shall be at right 

 angles to A B. 



Because D C = E C, 

 and that F C is common 

 to the two triangles DCF, 

 EOF; the two sides DC, 

 C F are = the two EC, C F, each to each, and the base 

 D F = base E F ; . . the angle D C F = angle 



Pr. 8. E C F,* and they are adjacent angles. But 

 when the adjacent angles which one straight line makes 

 with another are equal, each is called a right angle ;\ 



i. . . from the given point C in the given straight 

 line A B, a straight line C F has been drawn at right 

 any!' t to A B. IVTiich wan to be done. 



PROPOSITION XII. PROBLEM. 



To draw a straight line perpendicular to a given straight line 

 (AB) ofunlindted length, from a given point (C) without it. 



Take any point D upon the other side of A B, and 

 with centre C and radius C D describe the circle 



Port. 3. E G F' meeting A B in F, G : bisect F G in 

 tPr. 10. H,t and join C,H. 



The straight line C H shall be perpendicular to A B. 



cont. Draw CF, CG. Then because F H = G H, * 

 and H C common to 



the two triangles 

 FHC, OHO, the two 

 sides F H, H C are = 

 the two sides G H, 

 HC, each to each, and 

 the base C F = the 

 t Def. 12. baseCGjt 

 .-. the angle C H F = 

 the adjacent angle 



Pr. 8. C H G ; 

 each of them is .*. a 



t Def. 8. right angle, t and consequently CH is perpen- 

 dicular to A B ;t .'. from the given point C a 

 j-rf>r>irliciilar C B. to the given line AB has been drawn. 

 IVliich was to be done. 



PROPOSITION XIII. THEOREM. 



The angles which one straight line (AB) makes with 

 another (D C) upon one side of it, are either two right 

 angles, or are together equal to two right angles. 

 For the angle A B C is either equal to A B D, or it is not. 

 Def. 8. if ABC be = ABD, each is a right angle.* 



But if they are not equal, from B draw B E at right 

 t Pr- ll. angles to D C :t then the angles E BC, E }', U 



are right angles, also EBC = EBA+ABC. To 



each of these equals add EBD, .-. EBC + EBD = 

 -ABO* Again: ABD- 



VOL. i. 



E B D + E B A. Add A B C to each of these equals, .-. 



tAx.2. ABD + ABC = EBD + EBA4- ABC;t 



but it was demonstrated that E B C -f- E B D are equal 



to the latter three angles, .- ABD + ABC = EBD 



AX. i. -f E B C :* that is, the two angles ABD, 

 ABC are together equal to two right angles. 



Therefore the angles which one straight line, <tc. Q.E.D. 



COR. 1. It is manifest from this, that if two straight 

 lines cross one another, forming four angles at the point 

 of intersection, these four angles are together equal to 

 four right angles. 



COR. 2. And moreover, that if in a straight line any 

 point between its extremities be taken, from which any 

 number of straight lines 

 are drawn, some on one 

 side and some on the 

 other, all the angles thus 

 made at the point will 

 together be equal to four 

 right angles ; those on one 

 side of the proposed line 

 being together equal to 

 two right angles, and those on the other side equal also 

 to two right angles. 



PROPOSITION XIV. THEOREM. 

 If at a point (B) in a straight line (A B) two other 

 straight lines, (B C, BD), upon opposite sides of it, make 

 the adjacent angles (A B C, A B D) together equal to two 

 right angles, tlie two straight lines (B C, B D) shall be in 

 one and tlie same straight line. 



For, if B D be not in the same straight line with C B, 

 let BE be in the same straight line with it. Then be- 

 cause A B makes, with the straight line C B E upon one 

 side of it, the angles ABC, ABE, these are together = two 



Pr. 13. right angles.* 

 But ABC, ABD are 

 likewise together = two 



+ H JT>- right angles,t 

 .-. ABC + ABE = ABC 



AX. i. + A B D . * / 

 Take away the common 



angle ABC, and there 



+ AX. 3. remains ABE = ABD, the less to the greater, 

 which is impossible ; . . B E is not in the sariie straight 

 line with BC. And in like manner may it be proved 

 that no line, except BD, can be the prolongation of CB; 

 /. CB, BD are in one and the same straight line. Where- 

 fore, if at a point, <tc. Q. E. D. 



PROPOSITION XV. THEOREM. 



If two straight lines (A B, C D) cut one another, the 

 vertical (or opposite) angles shall be eqval ; that is, 

 AEC=DEB, andCEB = AED. 



The angles which A E makes with C D, on one side of 



Pr. 13. it, are together equal to two right angles ;* 

 that is, AEC+AED = two right angles. Again: 



the angles which D E 

 makes with A B are 

 also together equal 

 to two right nn- 

 t Pr. IS. gles ;* that 

 is DEA + DEB = two right angles ; . . AEC + AED 

 + AX. l. .-DEA+DEB.t Take away the common 



AX. 3. ang l e AED, and there remains AEC = DEB. * 

 And in a similar manner it may be proved that 

 C E B = A B D. Therefore, if two strnight lines, &c. 

 Q.E.D. 



PROPOSITION XVI THEOREM. 



If one tide (B C) of a triangle be prolonged, the exterior 

 angle (A C D) shall be greater than either of the 

 interior opposite (or more remote) angles. (B A C, 

 ABC). 



Pr. 10. Bisect A C in E,* draw B E, which prolong, 

 + Pr. 3. and make E F equal B E :t join F C. 



coiut Because A E = C E,* and E B = E F ; * A E, 

 E B are C E, E F, each to each and the angle A E B 



4 A 



