6. In Fig. 6.44, 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 = ½ ∠QPR.
Solution: Consider the ΔPQR. ∠PRS is the exterior angle and ∠QPR and ∠PQR are interior angles. So, ∠PRS = ∠QPR+∠PQR (According to triangle property) Or, ∠PRS -∠PQR = ∠QPR ———–(i) Now, consider the ΔQRT, ∠TRS = ∠TQR+∠QTR Or, ∠QTR = ∠TRS-∠TQR We know that QT and RT bisect ∠PQR and ∠PRS respectively. So, ∠PRS = 2 ∠TRS and ∠PQR = 2∠TQR Now, ∠QTR = ½ ∠PRS – ½∠PQR Or, ∠QTR = ½ (∠PRS -∠PQR) From (i) we know that ∠PRS -∠PQR = ∠QPR So, ∠QTR = ½ ∠QPR (hence proved).
