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.
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