Question

In the given figure $\angle 1=\angle 2$ and $\Delta NSQ\cong \Delta MTR$, then prove that $\Delta PTS\sim \Delta PRQ$

Answer 1

$\angle SQN=\angle TRM$ (CPCT as $\Delta NSQ\cong \Delta MTR$)

Since, $\angle P+\angle 1+\angle 2=\angle P+\angle PQR+\angle PRQ$(Angle sum property)

$\angle 1+\angle 2=\angle PQR+\angle PRQ$

$2\angle 1=2\angle PQR(as\angle 1=\angle 2and\angle PQR=\angle PRQ)$

$\angle 1=\angle PQR$

Also, $\angle 2=\angle PRQ$

And $\angle SPT=\angle QPR$ (common)

$\Delta PTS\Delta \sim \Delta PRQ$ (By AAA similarity criterion)