Given: \(\frac{QT}{PR}\)= \(\frac{QR}{QS}\)
∠1 = ∠2
R.T.P : △PQS ~ △TQR
Proof: In △PQR; ∠1 = ∠2
Thus, PQ = PR [∵ sides opp. to equal angles are equal]
\(\frac{QT}{PR}\)=\(\frac{QR}{QS}\) ⇒ \(\frac{QT}{PQ}\)= \(\frac{QR}{QS}\)
i.e., the line PS divides the two sides QT and QR of △TQR in the same ratio.
Hence, PS // TR.
[∵ If a line join of any two points on any two sides of triangle divides the two sides in the same ratio, then the line is parallel to the third side]
Hence, PS // TR (converse of B.P.T)
Now in △PQS and △TQR
∠QPS = ∠QTR
[∵ ∠P, ∠T are corresponding angles for PS // TR]
∠QSP = ∠QRT
[∵ ∠S, ∠R are corresponding angles for PS // TR]
∠Q = ∠Q (common)
∴ △PQS ~ △TQR (by AAA similarity)
