Question

In 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(PY2+XR2)=13PR2.

Answer 1

(1) PQ² + QR² = PR²
(2) XQ² + QR² = XR²
(3) PQ² + 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