BOOK m. PROP, xv. xix.] MATHEMATICS. PLANE GEOMETRY. 



667 



fron\_ the centre ; and chorda equally distant from the 

 centre are equal. 



. 1 in. Take E the centre of the circle A B D C,* and 

 from E draw E F, E G perpen- 



+ 12 I. diculars to A B, C D ; t 

 then if A B = C D, E F must be = 

 EG. For draw E A, E C. Then 

 because E F, from the centre, cuts 

 A B, a chord not through the centre, 

 at right angles, it bisects it,* .'. 



s in. A F = F B, .-. A B is 

 double of A F. For a like reason 

 C I) is double of C G. But A B 



+ Hyp. = C D,t .'. A F = C G. And because 

 EA=- EC, .-.EA* = EC 2 : but AF" + FE" = E A 2 , 



47 I. because A F E is a right angle ; * and, for a 

 like reason, C G 2 + G E = E C" .-. AF* + FE 2 = 

 CG J + GE ! : but AF J = CG J , because A F = C G, .-. 

 F & = G E 2 , . . F E = G E ; but chords of a circle are 

 said to be equally distant from the centre when the per- 

 pendiculars drawn to them from the centre are equal, t 



+ r>ef. s IIL . '. A B, C D are equally distant from the 

 centre. 



Next, let the chords A B, C D be equally distant from 

 the centre : that is, let the perpendiculars E F, E G be 

 equal : then shall A B = C D . For, the same construction 

 being made, it may, as before, be demonstrated that A B 

 is double of A F, and C D double of C G, and that A F" 

 -f F E- = C G : + G E J : but F E- = G E s , because F E 



Hyp. = GE,*.'. AF 1 =CG > , .'. AF = CG:but 

 A B, C D are the doubles of A F, C G, as already proved, 

 .'. A B = C D ; .'. equal chords, <tc. Q. E. D. 



PROPOSITION XV. THEOREM. 



The diameter is the greatest chord in a circle : and of all 

 others, that which is nearer to the centre it always 

 greater than one more remote : and the greater is nearer 

 to the centre than the less. 



Let E be the centre of the circle A B C D, and let B C 

 be nearer to it than F G : the diameter A 1) it the greatest 

 chord, and B C u greater than F G. 



Draw E B. E C. Then because E A - E B and E D 

 -EC, .'. AD = EB + EC: but E B + E C are greater 



JO I. than B C,* .'. A D it greater than B C. Again, 



draw EH, E K perpendiculars to 

 B C, F G ; then because B C is 

 nearer to the centre than F G,t 

 t Hyp K H is less than KK; 

 and, as was demonstrated in the 

 preceding proposition, E H 2 + H B 1 

 = EK3 + KF": but E H is less 

 than E K 3 , because E H is less than 

 E K, .'. H B* is greater than K F; 

 but BC is the double of H B,* 

 i in. and F G the double of K F. 

 .'. B C u greater than F G. 



Next, let B C be greater than F G, then B C shaU be 

 nearer to the centre than F G ; that is, the same con- 

 struction being made, E H shall be less than E K. Be- 

 cause B C is greater than F G, B H is greater than F K, 

 .-. 15 H is greater than F K. And B H 3 + H E 1 = 

 K K' + K E', of which B H 2 is greater than F K, /. 

 3 H" iUsi than EK", . . E H is less than EK ; that is, 

 B C u nearer to the centre than F G ; .'. the diameter, 

 <tc. Q. E. D. 



PROPOSITION XVL THEOREM. 



The straight line drawn at right angles to the diameter 

 (A B) of a circle (A B C), from 

 the extremity of it (A) falls with- 

 out the circle ; and no straight 

 line can be drawn from that ex- 

 tremity between the straight line 



A) ./ >B and the circumference, so as not 



to cut the circle. 



For if A E do not fall without 

 the circle, let a part of it, as A C, 

 fall withjn the circle ; and from 



the centre D draw D C, to the point C where A E meets 



the circumference. Then because I) A=D C, the angle 



+ 51. D AC = DCA;f but D A C is a right angle, * 



Hyp. .*. DC A is a right angle, .'. DAC+DCA 



+ 171. = two right angles, which is impossible,t .'. 



a part of A E, at right angles to B A, does not fall within 



the circle. And it may be demonstrated, in the same 



manner, that A E does not fall upon the circumference ; 



.'. A E falls without the circle. 



Moreover, between A E and the circumference no 

 t straight line can be drawn from 

 I A which does not cut the circle. 

 For let A F be supposed to be 

 drawn between them ; from D 

 draw D G perpendicular to A F, * 

 13 1. and let it meet the cir- 

 * cumference in H. Because AGO 

 is a right angle, and DAG less 

 + 17 1. than a right angle, t D A 

 19 1. is greater than D G ;* 

 but I) A =D H, .'. D H is greater 

 than I) G, a part than the whole, which is impossible, 

 .*. no straight line can be drawn from A between A K and 

 the circumference ; .'. the straight line, <fec. Q. E. D. 



COR. From this it is manifest that the straight line, 

 at right angles to a diameter of a circle, from an ex- 

 tremity of that diameter, touches the circle ; and that 

 it touches it in one point only, because if it met the 

 + JIH. circle in two points, it would fall within it.f 

 Also it is evident that there can be but one straight line 

 which touches a circle in the same point. 



PROPOSITION XVII. PROBLEM. 



To draw a straight line from a given point, either without 

 or in the circumference of a circle (B C D), which shall 

 touch tlie circle. 



First let the given point be without the circle an A. 



I III. Find E the centre,* and draw A E, cutting the 

 circle in D ; and with centre E, and radius E A, describe 

 the circle A F G ; from D draw D F at right angles to 



+ 11 1. E A,t and draw E B F, A B. A B shall touch 

 the circle BCD. 



Because E is the centre of the circles BCD, A F G, 

 E A = E F and E D = E B, .'. the two sides A E, E B 

 are = the two F E, ED, each to each, and the angle E 

 is common to the two triangles 

 A E B, F E D, .-. the angle F. B A 



.41. -KDF;* but EDF is 

 a right angle, .'. E B A ia a right 

 angle, and E B is a radius ; but a 

 straight line from an extremity of 

 a diameter, at right angles to it, 



+ 16 in. Cor. touches the circle, t 



' . A B touches the circle, and it is 

 drawn from the given point A ; which was to be done. 



But if the given point be in the circumference, as at 

 D, draw 1) E to the centre E, and D F at right angles to 



16 III. Cor. D E ; D F touches the circle.* 



PROPOSITION XVIII. THEOREM. 



If a straight line touch a circle, the straight line drawn 

 from the centre to the point of contact, shall be perpen- 

 dicular to the line touching the circle. 



This proportion in really included in Prop. XVI. : it afflrmi that the, 

 diameter u perpendicular to the touching line at its extremity : 

 and Prop. XVI. shows that the perpendicular to the diameter at 

 its extremity is the touching line, and the only touching line at 

 that extremity. The next proposition is equally superfluous : the 

 only thing prored by either U this viz., that a line which it per- 

 pendicular to the diameter must hate the diameter perpendicular 

 to it. The two enunciations are retained here only to avoid inter- 

 ruption in the numbering of the propositions. 



PROPOSITION XIX. THEOREM. 



(fa straight line touch a circle, and from the point of contact a 

 straight line be drawn at right anglei to the touching line, th* 

 cfntre of the circle ihall be in that line. 



