Question

In the given figure, ∠PQR = ∠PRQ and QR × PR = QT × SQ. Show that ΔPQS ~ ΔTOR.

Answer 1

Given:

∠PQR = ∠PRQ and QR × PR = QT × SQ

From ∠PQR = ∠PRQ, in △PQR, the sides opposite equal angles are equal. Hence, PR = PQ.

Now given, QR × PR = QT × SQ

substitute PR = PQ:

QR × PQ = QT × SQ

So, `(PQ)/(QT) = (SQ)/(QR)`

Also, P lies on QT and S lies on QR, so the angle between PQ and QS equals the angle between QT and QR. Therefore, ∠PQS = ∠TQR.

Thus, in triangles △PQS and △TQR,

• `(PQ)/(QT) = (SQ)/(QR)`
• ∠PQS = ∠TQR

Hence, by SAS similarity, △PQS ∼ △TQR.