Given, ΔNSQ≅ΔMTR and ∠1=∠2
To prove ΔPTS∼ΔPRQ
Proof since, ΔNSQ≅ΔMTR
So, SQ = TR ….(i)
Also, ∠1=∠2⇒ PT=PS ……(ii)
[ since, sides opposite to equal angles are also equal]
From Eqs. (i) and (ii), PSSQ=PTTR
⇒ ST∥QR [by convenes of basic proportionally theoreum]
∴ ∠1=∠PQR
And ∠2=∠PRQ
In ΔPTS and ΔPRQ, [common angles]
∠P=∠P
∠1=∠PQR
∠2=∠PRQ
∴ ΔPTS∼ΔPRQ [by AAA similarity criterion]