In the given figure, the side QR of ΔPQR is produced to a point S. If

1.	In the given figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR.
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

In the given figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR.

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