Question
In figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that
Solution
∵ ∠TRS is an exterior angle of ∆TQR ∴ ∠TRS = ∠TQR + ∠QTR ...(1)
| ∵ The exterior angle is equal to sum of its two interior opposite angles
∵ ∠ PRS is an exterior angle of ∆PQR
∴ ∠PRS = ∠PQR + ∠QPR ...(2)
| ∵ The exterior angle is equal to the sum of its two interior opposite angles
⇒ 2 ∠TRS = 2∠TQR + ∠QPR
| ∵ QT is the bisector of ∠PQR and RT is the bisector of ∠PRS
⇒ 2(∠TRS - ∠TQR) = ∠QPR ...(3)
From (1),
∠TRS - ∠TQR = ∠QTR ...(4)
From (3) and (4), we obtain
