Question

In the figure `(QR)/(QS) = (QT)/(PR)` and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.

Answer 1

In ΔPQR,

∠PQR = ∠PRQ

∴ PQ = PR   .......(i)

Given,

`(QR)/(QS) = (QT)/(PR)`  ........Using equation (i), we get

`(QR)/(QS) = (QT)/(QP)` ........(ii)

In ΔPQS and ΔTQR, by equation (ii)

`(QR)/(QS) = (QT)/(QP)`

∠Q = ∠Q

∴ ΔPQS ~ ΔTQR   .......[By SAS similarity criterion]