In a triangle PQR right angled at Q if QS = SR then prove that PR² = 4PS²- PQ²

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Solution

Given in right triangle PQR, QS = SR
By Pythagoras theorem, we have
PR² = PQ² + QR² → (1)
In right triangle PQS, we have
PS² = PQ² + QS²
= PQ² + (QR/2)² [Since QS = SR = 1/2 (QR)]
= PQ² + (QR²/4)
4PS² =4PQ² + QR²
∴ QR² = 4PS² −4PQ² → (2)
Put (2) in (1), we get
PR² = PQ² + (4PS² −4PQ²)
∴ PR² = 4PS² −3PQ²

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