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In Fig., if angle1 = angle 2 and $\Delta$ NSQ $\cong$ $\Delta$ MTR, then prove that $\Delta$ PTS $\sim$ $\Delta$ PRQ.

Answers (1)

Given :

$\Delta NSQ \cong \Delta MTR \; and\; \angle 1 = \angle 2$

To prove:-

$\Delta PTS \sim \Delta PRQ$

Proof:-

$\text {It is given that }\Delta NSQ \cong \Delta MTR$

$\therefore SQ=TR\; \; \; \; \; ...(1)$

$\angle 1=\angle 2$

We know that sides opposite to equal angles are also equal

$\therefore PT=PS \; \; \; \; \; \; \; .....(1)$

from equation (1) and (2)

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

According to Converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side.

$\therefore ST \parallel QR$

$\angle 1=\angle PQR$ $\text{(corresponding angle)}$

$\angle 2=\angle PRQ$ $\text{ (corresponding angle)}$

In $\Delta PTS$ and $\Delta PRQ$

$\angle P=\angle P$ $\text{ (corresponding angle)}$

$\angle 1=\angle PQR$

$\angle 2=\angle PRQ$

We know that if the corresponding angles of two triangles are equal, then the triangles are similar by AAA similarity criterion

$\therefore \Delta PTS \sim \Delta PRQ$ $\text{ (by AAA similarity criterion) }$

Hence proved

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