BOOK m. PROP, iv. vm.] MATHEMATICS. PLANE GEOMETRY. 



505 





, HyP . EA, EB. Then because AF = FB,t and FE 

 common to the two triangles AFE, BFE, and also 



.81. E A = E B, . . the angle A F E = B F E,* 



+ Def. 81- .'. each is a right anglerf .' . CD, drawn 

 through the centre, bisecting the chord 

 A B, that does not pass through the 

 centre, cuts it at right angles. 



Now let C D cut A B at right an- 

 gles : C D shall bisect A B. For, 

 the same construction being made, 

 because E A= E B, the angle E A F 



si. =EBF;* and the right 

 angle AFE = BFE, .-. in the two 

 triangles, EAF, EBF, there are 



two angles in the one = two in the other, each to each ; 

 and E F, opposite to equal angles, one in each, is com- 

 4 26 1. mon to both, .. A F = F B ;t that is, the chord 

 A B is bisected by the diameter C D, which it perpendicu- 

 lar to it ; .-.if a straight line, ic. Q. E. D. 



PROPOSITION IV. THEOREM. 



If in a circle (A B C D) two chords (A C, B D) cut one 

 another, and do not pass through the centre, they do not 

 bisect each other. 

 For let them cut in E ; and suppose A E = E C, and 



B E = E D. If one of the chords pass through the 



centre, it is plain it cannot be bisected by the other 



which does not pass through the centre, because the 



mi' Idle of the former is the centre. 



But if neither pass through the 



centre, take F, the centre;* 



ini. and draw F E. Then, 

 because F E, through the centre, 

 bisects A C not through the centre, 



. tin. it cms it at right angles, t 

 .'. F E A is a right angle. A'.'ain, 



Hyp. because FE bisects BD, 



a chord not through the centre, it cuts it also at right 

 + s in. angles, + .'. FEB is a right angle; but it was 

 proved that F E A is a right angle, .'. F E A = F E B, 

 the less to the greater, which is impossible, .'. A C, B D 

 do not bisect one another ; .'. >f in a circle, ic. Q. E. D. 



PROPOSITION V. THEOREM. 



If two circlet (A B C, C D O) cut one another (as in the 



point C), thfy shall not have the tame centre. 

 For, if it be supposed possible, let E be the common 

 centre. Draw E C, and draw 

 E F G, meeting the circumfer- 

 ences in F and G. Then, because 

 E is the centre of the circle ABC, 

 Def. is i. EC = EF.* Again, 

 because E is the centre of the 

 circle C D G, E C - E G. But 

 it was shown that E C = E F, .'. 

 E F = E G, the less to the 

 greater, which is impossible, .'. 

 the two circles have not a common centre ; .'. if two circles. 

 Ac. Q. E. D. 



PROPOSITION VI. THEOREM. 



If two circlet (A B C, C D E) touch one another, they shall 

 not have the same centre. 



Let them touch at C, then if they 

 have a common centre, let it be F. 

 Draw F C, and F E B, meeting the 

 circumferences in E and B. Then, 

 because F is the centre of ABC, 

 FC = FB;*also, be- 

 cause F is the centre 

 of CDE,FC = FE:* but it was 

 shown that F C = F B, .-. FE = 

 FB, the less to the greater, which is impossible, .-. the 

 two circles have not a common centre ; ,', if two circles, <tc. 

 Q.E.D. 



PROPOSITION VII. THEOREM. 

 If any point (F), not the centre, be taken in the diameter 



Def. 12 I. 



(A D) of a circle (A B CD), then of all lines (F A, F B, 

 F C, <fcc.) which can be drawn from it to the circum- 

 ference, the greatest is that (F A) which passes through 

 the centre (E), and (F D), the other part of the diameter, 

 is the least. And of the others, that which is nearer to 

 the line through the centre is always greater than one 

 more remote. And from the same point (F) there can be 

 drawn two, and only two, straight lines that are equal, 

 one upon each side of the diameter. 



Draw E B ; then because any two sides of a triangle 

 are together greater than the third side, F E, E B are 

 20 1. greater than F B ;* but E A = E B, .-. F E + 

 E A ; that is, F A is greater than F B, and F B is any 

 other line, . . the line through the centre is the greatest. 

 Again, draw E C : then, because E B = E C, and F E 

 common to the triangles B E F, C E F, . . the two sides 

 F E, E B are = the two F E, E C, each to each ; but the 

 angle B E F is greater than C E F, . . F B is greater than 

 t 24 I. FC,t .' the line 

 nearer to that through the centre 

 is greater than one more re- 

 mote. 



Again, draw E G : then be- 

 cause E F + F G are greater 

 20 1. than E G,* and that 

 EG=ED, .-.EF + FGare 

 greater than E D. Take away 

 the common part E F, /. F G 

 is greater than F D, and F G 

 is any other line, .*. F D is the 

 least of all the lines from to the circumference. 



Also two, but only two, equal straight lines can be 

 drawn from F to the circumference, cue upon each side 

 of the diameter AD. For FG being any straight line 

 iM i from F, at E make the angle F E H = F E G,t 

 and draw F H. Then, because EG= EH, and EF 

 common to the triangles G E F, *H E F, the two sides 

 E G, E F are = the two EH, E F, each to each, and the 



con*. angleGEF = HEF,*.-.FG = FH.t But, 

 + 41. besides F H, no other line equal to F G can be 



drawn from F to the circumference. For, if there can, 

 let it be FK: then, because FK =FGandFG = FH, 

 . . F K = F H ; but one of these must be nearer to the 

 line through the centre than the other, so that a line 

 nearer to that through the centre is equal to one more 

 remote, which has been proved to be impossible, . . besides 

 F H no other line = F G can be drawn ; . . if any point, 

 <tc. Q. E. D. 



PROPOSITION VIII. THEOREM. 



If any point (D) be taken without a circle (A B F), and 

 straight lines (D A, D E, D F, <tc.) be drawn from it to 

 the circumference, of which one (DA) passes through the 

 centre (M), of those which fall upon the concave part of 

 the circumference, the greatest is that (DA) which 

 pusses through the centre ; and of the rest, that which is 

 nearer to tiie one through the centre is always greater 

 than one more remote. But of those which fall upon 

 the convex part of the circumference, the least is that 

 (D G) which, when prolonged, passes through the centre ; 

 and of the rest, that which is nearer to the least is 

 always less than one more remote. Also two, and only 

 two, equal straight lines can be drawn from the same 

 point (D) to the circumference, one upon each side of the 

 line through the centre. 



i ill. Take M the centre of the circle,* and draw 

 M E. Then because A M = E M, add M D to each, . . 

 AD = EM+MD; but E M + M D are greater than 



t 20 I. E D,f .'. A D is greater than E D, and E D is 

 any other line, .'. A D, the line through the centre, is the 

 longest. 



Again, draw M F. Then, because E M = F M, the 

 two EM, M D are = the two F M, M D, each to each ; 

 but the angle E M D is greater than F M D, /. E D is 



24 I. greater than F D ; that is, a line D E, nearer to 

 that through the centre, is greater than D F, one more 

 remote. 



Now draw M K. Then because M K + K D are greater 



