360 Mr. Walker on some Geometrical Principles [May, 



in n. Lastly, from the centre c, and parallel to A E, draw c Y 

 meeting E T in Y ; which point we shall prove to be in the peri- 

 phery of the circle. 



From the similarity of the triangles c m Y, A«E, cm : A m 

 :: c Y : A E. But we shall prove that cm : A m :: A c : A E ; 

 and .*. that A c = c Y. For the triangles c m E, m n D, being 



similar, c m : m n :: c E : D n ; that is, c m : :: c E : — 



/. c m : A m :: c E : E e. But c E : E e :: A c : A e on ac- 

 count of the bisection of the angle y A c by the line A E]. .•. cm 

 : A m :: A c : (A e) A E. But we have before seen that c m : 

 A m :: c Y : A E. .*. c Y = A c, and the point Y is in the peri- 

 phery. But the angle b c Y being equal to b E A, that is, to the 

 third part of A B a, the line E T Y trisects the arc A b a in Y. 

 Q. E. D. 



Cor. 1. (Fig. 6.) — If from p, the point where A c meets a E, 

 the line p d be drawn to the middle point of A E, and be pro- 

 duced on each side to meet the periphery in a and q ; the line 

 <r dp q is the side of an equilateral triangle inscribed in the circle 

 ABoi; and if the lines E ce, E q be produced to meet the 

 circle A E a e in m and n, the triangle E m n is an equilateral 

 triangle inscribed in that circle. 



For the triangle A p E is isosceles [for angle p A~E — c A~B 

 + BAE = BEA + 2BAE=2BEA = AE|)] : .:p d is 

 perpendicular to A E, and .*. bisects the parallel radius c x per- 

 pendicularly. .•. a q is the side of an equilateral triangle, of 

 which Y o is the altitude. Now drawing the line A e, it must 



pass through y [foro A e = a E e = — - — = a A y~[. .•. Ay is 



parallel to p d; and .'. E y, the hypotenuse of the right-angled 

 triangle E A y, is bisected by p d. But it is also bisected by 

 Y x. .'. E y and & q bisect each other in o. .*. tn E q y is a 

 parallelogram. .*. the angles «Ef and <j(Ec are each of them 

 30° [for ffEf = (cE;/ — y E 6 = qy E — x y E = qy x = 

 30°, and in like manner, q~Ec = q E 3/ +yE J = «^ E + 

 a'3/ E = q y x = 30°]. .*. raE n is an equilateral triangle 

 inscribed in the circle AEa«. 



Cor. 2. — The arc B a is the third part of A « B, and B q the 

 third part of A a B. For <eY q being an eauilateral triangle, 

 arc a, Y = 2 «. x. But A Y = 2B,r. .-. a Y - A Y = 2 <e B. 

 .-. &c. In like manner 5 Y + A Y = A Y q = 2 B q v &c. 



Cor. 3. — If from c be drawn a parallel to A a, meeting A Ein 

 k, and Y E in /, Y / is the third part of Y E. For the triangles 

 ex. E and c z k are each of them isosceles. /. k E = diameter. 

 But from the similarity of the triangles klE, c IY, Y I : I E :: . 

 c Y : k E :: 1 : 2. .-. Sec. It is plain (hat vz must cut off another 

 third part of E Y. Also, that c k= Y y. 



Cor. 4. — Producing A c and A B to meet the circle A E a e in 

 r and s, the chord E r is equal toAo; and the semicircular arc 



