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.

Answer 1

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.