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

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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.

Updated on 10-Oct-2022 13:28:02