Given,
ΔNSQ≅ΔMTR and ∠1=∠2 To prove that
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 converse of basic proportionally theorem]
∴ ∠1=∠PQR and
∠2=∠PRQ In
ΔPTS and ΔPRQ,
∠P=∠P [common angles]
∠1=∠PQR ∠2=∠PRQ ∴ ΔPTS∼ΔPRQ [by AAA similarity criterion]