In figure below, if \( \angle 1=\angle 2 \) and \( \triangle \mathrm{NSQ} \cong \triangle \mathrm{MTR} \), then prove that \( \triangle \mathrm{PTS} \sim \triangle \mathrm{PRQ} . \)
![](/assets/questions/media/153848-1651315434.png)
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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.
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