Question

In the following figure, `("QR")/("QS") = ("QT")/("PR")` and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Answer 1

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")`

∠Q = ∠Q = 1

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