1824.] connected with the Trisection of an Arc. - 361 



t E u is bisected in s. For angle A E a = E A r, the triangle 

 A p E being isosceles. .-. A a = Er. But arc E t = — -, and 



arc Es = — . .•. arc £ s = — — = quadrant. 



Cor. 5. (Fig. 7.) — If By'be drawn parallel to A E, it is equal 

 to E c. For drawing^/'g parallel to the axis, and meeting A E 

 in g,fg = g Z, on account of the equality of the angles g Z/, 

 g/Z. .-.fg + B c, or Ec, = g Z + E Z = B/. 



Pkop. IV. (Fig. 7.) 



The same things being supposed, E r the segment of the tri- 

 secting line intercepted between the two circles is the third part 

 of EX, the segment of it included in the circle AE«e. 



Dem. — Draw A n, making the angle EA« = AEi, and 

 cutting E X in s. The triangles E A n, AE.r being identical, 

 A» = Ei=EX. But E X trisecting A a, and being parallel 



to a n, trisects An. .'.As being equal to — ^, s X = — . But 

 A s, and .'. s X, = s r, the angles at the base of the triangle 

 A s r being equal. [For angle A r s = ' = AEc. But it 



also equals A Er + E A r. .-. angle E A r = X E e. .*. angle 

 r A s, or E A n — E A ;•, = AEX-,rEe = A E e = A /• sj. 

 t? EX 



Cor. — Drawing a r v, r v = ",-. For the triangle a r X is simi- 



lar to A E « [for angle a r X = A E a, and angle rXa = EAa]. 

 V a r = r X, and .-. r v = E r. .-. &c. 



Scholium. — It is plain that A n bisects Ej/ in o ; since Ao = 

 E o, and the angle E A y is a right angle. 



Prop. V. (Fig. 7.) 



The same things being supposed, if there be drawn a diameter 

 of the circle A E a e perpendicular to the axis, and meeting the 

 circle A B a b in T, the line E T is a tangent, and equal to E /, 

 the line drawn from E to the point where z Z produced meets 

 the circle A E a e. 



For a Z passes through k, the centre of the circle AEae; and 

 the triangle Z k c is isosceles, and similar to the triangle E Z c. 

 .-. c E : cZ::cZ:ck; that is, c E : c T :: r T : c ft. ^ .-. the 

 angle c T E is a right angle, and E T a tangent. .-. E T- = 



(r E . E k) = (^ . 2 E ft) = (E d .Ee) = Et*. .: E T = 



E /. 



Cor. — Hence it appears, that if c T E be a right-angled trian- 

 gle, in which from the right angle the perpendicular T ft is let 



