In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
In ΔPQR, ∠PQR = ∠PRQ
∴ PQ = PR ...(i)
Given,QR/QS = QT/PR
Using (i), we get
QR/QS = QT/QP ...(ii)
In ΔPQS and ΔTQR,
QR/QS = QT/QP [using (ii)]
∠Q = ∠Q
∴ ΔPQS ~ ΔTQR [SAS similarity criterion]
