Question

below is the question

Answer 1

Answer:

In ΔQTR, ∠TRS is an exterior angle.

∠QTR + ∠TQR = ∠TRS

∠QTR = ∠TRS − ∠TQR (1) For ΔPQR, ∠PRS is an external angle.

∠QPR + ∠PQR = ∠PRS ∠QPR + 2∠TQR = 2∠TRS (As QT and RT are angle bisectors)

∠QPR = 2(∠TRS − ∠TQR) ∠QPR = 2∠QTR [By using equation (1)]

∠QTR = 1/2 ∠QPR.

hence proved.