670 



MATHEMATICS. PLANK GEOMETRY. [BOOK HI. mop. XXXIT. XXXVIL 



Hut if be not a right 

 angle, make B A D - C,* 



. m> mid draw A K at 



t u I. right angle* to AD.f 



>' Bisect A Bin F, 

 and draw F O at right angles 



ni. to A B :t FG will 

 meet A F. in some poiut O, 

 beoauM O A F is lev than a 



AX ll. right angle ; and 

 if from O as centre, with 



radios O A, a circle be described, it will pass through B, 

 and the segment A II B will be that required. For draw 

 Q B : then because A F - F B, the two sides A F, KG 

 are the two B F, F G, each to each ; and the angle 

 rot. AFG -BFG.t G A- GB;' .-.the 



41. circle, with centre G, and radius G A, must 

 pass through B. Again : because from A, an extremity 

 of the diameter A E, A 1) 



is drawn at right angles 

 to A E, . . AD touches 

 t is in. Cor. the circle :f 

 and because A B, drawn 

 from the point of contact, 

 cuts the circle, the angle 

 D A B = H, in the alter- 



31 in. nate segment :* 

 butL) AB - C, .-. H = 



C ; .-. upon A B the segment A H B of a circle is described, 

 which contains an angle II = C. Which was to be done. 



PROPOSITION XXXIV. PROBLEM. 



From a given circle (A B C) to cut off a segment, which 

 shall contain an angle equal to a given angle (D). 



Take any point B in the circumference, and draw E F 

 touching the circle in B,* 



17 in. and make the /^ \ 

 + 231. angle FBC=D.f . / \ 



The segment B A C shall 

 contain an angle A = U. 

 Because E F touches the 

 circle at B, the angle FHC 

 B A C, in the alternate 



si in. segment:* but 



t Con.t. FBC = D,f 

 .'. the angle A, in the segment B A C, ia = D ; . . from the 

 given circle the segment B A C ia cut off containing an 

 angle = D. Which was to be done. 



PROPOSITION XXXV. THEOREM. 



Vtico chords (A C, B D) in a circle, cut 

 one another, the rectangle (AE'KC) 

 contained by the segments of one of 

 thrm, is equal to the rectangle 

 (UK-ED) contained by the segments of 

 the other. 



If the chords intersect in the cen- 

 tre E, it is evident that AE, EC, 

 B E. ED, being all equal, the rect- 

 angle AE EC BE ED. 



But if one of them, B D, pass through the centre, and 

 cut the other A C, which does not pass through the cen- 

 tro, at right angles, in E, then, if B D be bisected in F, 

 F is the centre of the circle ; join A, F. Then, because 

 B D, through the centre cuts A C, 

 not through the centre, at right 



3 in. angles in E, A E = E C.' 

 And because BD is cut into two 

 equal parts in F, and into two un- 

 equal parts in E, B E E D + E F 1 



II - F B 1 , or F A* ;t but 



471. AE'-(-EF 1 =FA 1 .-. 

 B E E D + E F* - A E' -f E F. 

 Take away the common square 

 EF 1 , .-. BE ED- AE'-AE 



E C. Again, let H 1), which posses through the centre, 

 cut A C, which does not pass through the centre in E, 

 but not at right angles; then, as before, if B D bo bisected 



in F, F is the centre of the circle. 

 Draw A F, and from F draw F G 



III. perpendicular to At',* .-. 

 tllll. AG-GC',t .-. AEEC + 



6H. EG-'=AG.* To each of 

 these add G F 1 , .- . A E-E C + E G 

 + G F - A G 1 + G F ; but E G 



+ 471. + G F* - E F,t and A G 

 + G F= A F', .-. A E-E C + EF=> 



5 II. A F - F B ; but F B = B E-E D + E F 1 ,* 

 . . A E-E C -f E F* - B E-E D + E F. Take away 

 EF, .-. AE-EC -BE-ED. 



Lastly, let neither of the chords A C, B D pass through 

 the centre. Take the centre F,t 

 Mill, and through E, the in- 

 tersection of the chords, draw the 

 diameter G E F H. Then, because 

 AEEC = GEEH, as also BE-ED 

 GE-EH, as already proved, .% 

 AE-EC = BE-ED; .'. if two 

 chords, <fec. Q. E. D. 



PROPOSITION XXXVI. 



THEOREM. 



If from any point (D) without a circle (A B C) two straight 

 lines be drawn, one of which (D A) cuts the circle, and 

 the other (D B) touches it, the rectangle (A D'D C) con- 

 tained by the whole line cutting the circle, and the part 

 imthout the circle, shall be cqiial to (D B 2 ) the sqttare of 

 the line which touches U ; that is, A D -D C = D B 2 . 

 The line D C A either passes through the centre, or it 

 does not ; first let it pass through 

 the centre E ; and draw E B ; then 

 + is in. B is a right angle ;f and 

 because A C is bisected in E, and 

 prolonged to D, A D-D C + E C 1 = 



BUI. ED';* butEC = EB, 

 /. AD-D C -f E B 2 = E D J ; but 

 E D 8 = E B 5 -j- B D ! , because B is a 



+ 471. right angle,f .'. AD'D C 

 + E B"= E B ! + B D 2 . Take away 

 EB 3 , .-. ADDC=.BD 2 . But if 

 DCA do not pass through the centre, 



1III. take the centre E,* and 

 draw E F perpendicular to A C ;t 



+ U I. draw also E B, E C, ED. Then because E F, 

 passing through the centre, cuts AC, 

 not through the centre at right 



3 in. angles, it also bisects it,* 

 . . A F = F C ; and because A C is 

 bisected in F, and prolonged to D, 



ten. ADDC + FC S = FD 2 .t 

 To each of these add FE 2 , .-. 

 ADDC + FC 2 +FE 2 = FD 2 + 

 FE 2 ; but ED 2 = FD 2 + FE , 

 because E F D is a right angle ' 



471. and, for the same reason, 

 EC 2 = FC" + FE 2 , .-. AD-DC 

 + EC 2 = ED ; but E C = E B, 

 . . A D-D C + E B* - E D 2 ; but 



EB' + BD S = ED 2 , because B is a right angli-,t 

 + 471 .-. ADDC + EB 2 = EB 2 +BD. Take 

 away E B 2 , .'. A D D C = BD*. 



COR. If from any point without a circle, two straight 

 lines cutting it be drawn, the rectangles of the whole 

 lines and the parts without the circle are equal ; for each 

 rectangle is equal to the square of the line drawn from 

 he po int to touch the circle. 



PROPOSITION XXXVII. THEOREM. 



Jf from a point (D) without a circle (ABC) there be 

 drawn two straight lines, one (D C A) to cut the circle, 

 and the other (DB) to meet it ; and if the rect<iii<jlr. 

 (A D D C) of the whole cut ting line and the part with- 

 out the circle be equal to (D B 2 ) the square of the line 

 which meets it, the line (D B) which meets shall touch 

 the circle. 

 17 in. Praw D E touching the circle in E :* find 



