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Question
In the given figure, S is a point on side QR of ΔPQR such that ∠QPR = ∠PSR. Use this information to prove that PR2 = QR × SR.
Solution
Given: ∠QPR = ∠PSR
In ΔQPR and ΔPSR,
∠QRP ≅ ∠PRS ......[Common angle]
∠QPR ≅ ∠PSR ......[Given]
So, according to the AA similarity criterion,
ΔQPR ∼ ΔPSR
∴ `(QR)/(PR) = (PR)/(SR)` .......[C.S.S.T.]
⇒ QR × SR = PR2
Hence proved.
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