In figure, if `angle1=angle2` and `DeltaNSQ=DeltaMTR`, then prove that `DeltaPTS~DeltaPRQ`.

Explanation

Given `DeltaNSQcongDeltaMTRandangle1=angle2`
to prove `DeltaPTS~DeltaPRQ`
Proof Since, `DeltaNSQcongDeltaMTR`
So, SQ=TR…..(i)
Also, `angle1=angle2rArr`PT=PS..(ii)
[since, sides opposite to equal angles are also equal]
From Eqs. (i) and (ii) `(PS)/(SQ)=(PT)/(TR)`
`rArr STabs()QR` [by convense of basic proportionality theorem]
`thereforeangle1=anglePQR`
and `angle2=anglePRQ`
in `DeltaPTS and Delta PRQ` [common angles]
`angleP=angleP`
`angle1=anglePQR`
`angle2=anglePRQ`
`thereforeDeltaPTS~DeltaPRQ` [by AAA similarity criterion]
Hence proved