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In Fig., if angle1 = angle 2 and \Delta NSQ \cong \DeltaMTR, then prove that \DeltaPTS \sim \DeltaPRQ.

 

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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