TO GEOMETRICAL PROBLEMS. 
259 
m 
PQ = PA x PB, 
or 
nx 
t x x H ~ab; there- 
m 
fore inx* + nx 1 — mab, 
and 
x - 
■mab 
m 4- n 
R 
Through the centre O, draw PAB ; put PA — a, 
PB — b, PR z= x, and let 
the given ratio of PR to 
RQ be that of m to n; 
then it will be, as m : n :: 
x : — RQ; therefore 
m 
PQ =r x -f : but PR x 
PROBLEM VIII. 
The sum of the two sides of an isosceles triangle ABC 
being equal to the sum o f the base and perpendicular, and 
the area of the triangle being given; to determine the 
sides. 
Put the semi-base AD — x, the perpendicular CD 
— y, and the given area 
ABC a a : so shall xy — 
a 1 , and 2\fxx + yy zz <2x 
+ y (by El. 47. 1. and the 
conditions of the problem.) 
Now, squaring both sides 
of the last equation, we 
have 4xx + 4yy zz 4xx + 
4xy + yy 5 whence 3yy zz 
4xy, and consequently y zz~~: which value, substituted 
in the former equation, gives —— — a*; from whence 
x — V — *a>/ 3; y {= —\U >/ 3 ; and AC