In given figure ∠1 = ∠2 and ∆NSQ ≅ ∆MTR, then prove that ∆PTS ~ ∆PRQ.

Given that: ∠1 = ∠2

and ∆NSQ ≅ ∆MTR

To prove: ∆PTS ∼ ∆PRQ

Proof: Since ∆NSQ ≅ ∆MTR,

⇒ ∠NQS = ∠MRT [∵ corresponding parts of congruent triangles are congruent]

⇒ ∠PQR = ∠PRQ …(i)

In ∆PTS,

∠P + ∠S + ∠T = 180° [∵ sum of angle of a triangle is 180°]

⇒ ∠P + ∠1 + ∠2 = 180°

⇒ ∠P + ∠1 + ∠1 = 180° [∵ given]

⇒ ∠P + 2∠1 = 180° …(ii)

In ∆PRQ,

∠P + ∠Q + ∠R = 180°

⇒ ∠P + ∠Q + ∠Q = 180° [from (i)]

⇒ ∠P + 2∠Q = 180° …(iii)

Comparing equations (ii) and (iii), we get

∠P + 2∠1 = ∠P + 2∠Q

⇒ ∠1 = ∠Q …(iv)

Thus, we have got

∠PST = ∠PQR [from equation (iv)]

∠SPT = ∠QPR [common angle]

∴ ∆PTS ∼ ∆PRQ by AA-similarity.