Question

In the given figure PS=TS, RP=RQ and QR is parallel to ST prove that TPQ =90°​

Answer 1

Given :  PS = TS, RP = RQ and QR is parallel to ST.

To Find : Prove that angle TPQ is a right angle.​

Solution:

RP = RQ

=> ∠QPR = ∠PQR = x

=> ∠QRS = x + x = 2x  ( exterior angle of triangle = Sum of opposite angles of triangle)

QR || ST

=> ∠QRS  = ∠TSR

=> ∠TSR = 2x

=>  ∠TSP = 2x

PS = TS

=> ∠TPS = ∠PTS

∠TPS + ∠PTS + ∠TSP = 180°

=> 2∠TPS + 2x = 180°

=> ∠TPS +  x = 90°

∠QPR  = x

=>  ∠TPS +  ∠QPR  = 90°

=>  ∠TPQ = 90°

QED

Hence proved that TPQ is a right angle.​

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