In figure below, if $ \angle 1=\angle 2 $ and $ \triangle \mathrm{NSQ} \cong \triangle \mathrm{MTR} $, then prove that $ \triangle \mathrm{PTS} \sim \triangle \mathrm{PRQ} . $
"

Given:

\( \angle 1=\angle 2 \) and \( \triangle \mathrm{NSQ} \cong \triangle \mathrm{MTR} \)

To do:

We have to prove that \( \triangle \mathrm{PTS} \sim \triangle \mathrm{PRQ} . \)

Solution:

$\triangle NSQ \cong \triangle$ MTR$

This implies,

$SQ = TR$......….(i)

\( \angle 1=\angle 2 \)

This implies,

$PT = PS$.......….(ii) (Sides opposite to equal angles are equal)

From (i) and (ii),

$\frac{PS}{SQ} = \frac{PT}{TR}$

Therefore, by converse of basic proportionality theorem,

$ST \| QR$

This implies,

$\angle 1 = PQR$

$\angle 2 = \angle PRQ$

In $\triangle PTS$ and $\triangle PRQ$

$\angle P = \angle P$               (Common)

$\angle 1 = \angle PQR$

$\angle 2 = \angle PRQ$

Therefore, by AAA similarity,

$\triangle PTS \sim \triangle PRQ$

Hence proved.

Tutorialspoint

Updated on: 10-Oct-2022

49 Views

Kickstart Your Career

Get certified by completing the course

Get Started