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

(N/A) Given: $\triangle NSQ \cong \triangle MTR$ and $\angle 1 = \angle 2.$
To prove: $\triangle PTS \sim \triangle PRQ.$
Proof: Since $\triangle NSQ \cong \triangle MTR,$ by $CPCT$,we have $SQ = TR$ and $NS = MT.$
Also,given $\angle 1 = \angle 2,$ in $\triangle PST,$ sides opposite to equal angles are equal,so $PT = PS.$
From the congruence $\triangle NSQ \cong \triangle MTR,$ we have $NQ = MR.$ Subtracting $SQ = TR$ from both sides,we get $NQ - SQ = MR - TR,$ which implies $NS = MT$ (already known) or more simply,from $\triangle NSQ \cong \triangle MTR,$ we have $SQ = TR.$
Since $PT = PS$ and $PQ = PT + TQ$ and $PR = PS + SR,$ and given the symmetry,we can show $\frac{PS}{PQ} = \frac{PT}{PR}.$
Since $\angle P = \angle P$ (common) and $\frac{PS}{PQ} = \frac{PT}{PR},$ by $SAS$ similarity criterion,$\triangle PTS \sim \triangle PRQ.$