Question

Q.No.9. The side QR of a triangle PQR is produced to a point S. If the bisector of
∆ PQR and ∆ PRS meet at point ∆ QPR (See given Figure), then prove that;
∆QTR=1/2∆QPR​

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

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