In given figure, is a right triangle PQR, right angled at Q. X and Y are the points on PQ and QR such that PX : XQ= 1 : 2 and QY : YR= 2 : 1. Prove that 9(PY 2+ XR 2 )= 13PR 2 . |
In triangle QXR,
XR^2 = (2x)^2 + (3x)^2
= 13x^2
In triangle PQY,
PY^2 = 13x^2
IN triangle PQR ,
PR^2= 18 x^2
now substitute the value of both in the equation
9( 13 x^2 + 13x^2 ) = 13 * 18x^2
234 x^2 = 234 x^2
since LHS =RHS
it is proved....