1.

In the given figure, \(\frac{QT}{PR}\)=\(\frac{QR}{QS}\) and ∠1 = ∠2. Prove that △PQS ~ △TQR.

Answer»

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)


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