358 Mr. Walker on some Geometrical Principles [May, 



a t :: 2 : ]. .•. A b, bisecting the angle EA«, cuts E e into 

 segments in the same ratio. 



Cor. 2. (Fig. 3.) — Drawing the diameter E p, and the line Ap, 

 the latter bisects the radius B c in q, and trisects B E in r. Let 

 jN be the point in which E P, perpendicular to E e, intersects 

 A p. The triangles ~E~N p,ey a are similar [for the angles at p 

 and a stand on equal arcs, as also the angles NEp and A ux ; 

 but A ax = aye']. .-. EN : E p.: y e : y a :: I : 2. vEN 

 = radius. .-. cq = semiradius. But the triangles ENr,Byr 

 are similar. .-. Er : r B :: EN : q B :: 2 : 1. .-. E B is trisected 

 in r. It is plain, that if from E we draw E g parallel to Ap, it 

 must bisect the radius b c in Q. 



Cor. 3. (Fig. 3.) — Drawing through Q an ordinate (which is 

 the base of an inscribed equilateral triangle), the lines Be, b g 

 intersect this ordinate in the same point o. For the arcs A E, 

 p g being equal, as also the arcs E b, p B, .-. the arc A b = Bg. 

 ••• b g is parallel to A B. .-. the triangle B h b is isosceles. 

 But its altitude h c is trisected by B e [as Be trisects a D]. 

 .'. the side b h is bisected by B e in o. [For universally, if from 

 an angle at the base of an isosceles triangle a line be drawn cut- 

 ting the altitude and leg, the segments of the leg are to each 

 other as the upper segment of the altitude to twice its lower 

 segment.] But the ordinate through Q, the middle point of b c, 

 must also bisect b h . .-. &c. 



Cor. 4. (Fig. 4.) — Let ibe the point in which the perpendicu- 

 lar from e meets a B. E? is parallel to A B. For drawing i I 

 parallel to A a, and cutting B E in r, this parallel is trisected in 

 r; and from the similarity of the triangles E I r, Bur, BE is 

 also trisected in r. .-. the triangles B r I, Er; are similar. 

 /.IB and E i are parallel. 



Scho/utm. — In the assigned solution of the preceding problem, 



the base of the given segment is always one of the extreme sides 



of the triangle ; and the other extreme is the chord of the arc 



intercepted by the trisecting lines. But if the given segment 



contain an angle of less than 60°, its base may be the middle 



side of an inscribed triangle, having its sides in arithmetical 



progression. And from what has been established, it is plain 



how we may construct that triangle. Let a x (fig. 3) be given as 



the middle side of such a triangle, in the segment a B x. Then 



we have e, the middle point of the arc ax. .-. we have P, the 



opposite extremity of the diameter e c P ; and .-. we have a P. 



But we have seen (Cor. 2) that a P bisects the radius B c in q. 



If, therefore, with the centre c, and the interval of the semiradius, 



we describe a circle, its occurse with a P determines the point 



<j. .-. we have the diameter B b, and .-.its ordinate a A, one of 



the extreme sides of the triangle ; the greatest or the least, 



according to the point of intersection of the described circle 



with the line a P, through which we draw the diameter. 



