ΔPQR, M∠Q=90, M ∈ QR.Prove that PR^2=PM^2+3RM^2
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Explanation: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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