Question

PQR is a right angle triangle right angled at q. X and Y are 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

Answer 1

Anyway, assuming that PQR is indeed right angled at Q, then by Pythagorean theorem, we get: (1) PQ² + QR² = PR²
(2) XQ² + QR² = XR²
(3) PQ²
​PQ+ QY² = PY² Since PX:XQ = 1:2 ---> PX = 1/3 PQ, XQ = 2/3 PQ Since QY:YR = 2:1 ---> QY = 2/3 QR, YR = 1/3 QR Plugging into (2) we get (2/3 PQ)² + QR² = XR² 4/9 PQ² + QR² = XR² 4 PQ² + 9QR² = 9XR² .... (4) Plugging into (3) we get PQ² + (2/3 QR)² = PY² PQ² + 4/9 QR² = PY² 9PQ² + 4 QR² = 9PY² .... (5) Adding (4) and (5) we get 9XR² + 9PY² = 4PQ² + 9QR² + 9PQ² + 4QR² 9 (XR² + PY²) = 13 (PQ² + QR²) 9 (XR² + PY²) = 13 PR²