Question

In the following figure, \(\frac{QR}{QS}=\frac{QT}{PR}\) and \(\angle 1 =\angle 2\)

Show that ΔPQR∼ΔTQR

Answer 1

Given: \(\frac{QR}{QS}=\frac{QT}{PR}\) and \(\angle 1 =\angle 2\)

To Prove: ΔPQR∼ΔTQR

Proof: In ∆PQR,
\(\angle\)PQR = \(\angle\)PRQ
∴ PQ = PR ………………(i)

Using (i) we obtain
\(\frac{QR}{QS}=\frac{QT}{QP}\).............(ii)

In ΔPQS and ΔTQR,
\(\frac{QR}{QS}=\frac{QT}{QP}\) [using (ii)]
\(\angle\)Q=\(\angle\)Q
∴ ΔPQS∼ΔTQR

Hence Proved