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 =1/2 ∠QPR.

It is given in the question that: Bisectors of ∠PQR and ∠PRS meet at point T

To prove,

∠QTR = ½ ∠QPR

Proof:

In ΔQTR,

∠TRS = ∠TQR + ∠QTR (Exterior angle of a triangle equals to the sum of the two opposite interior angles)

∠QTR = ∠TRS - ∠TQR -----------(i)

Similarly in ΔQPR,

∠SRP = ∠QPR + ∠PQR

2∠TRS = ∠QPR + 2∠TQR  (∵ ∠TRS and ∠TQR are the bisectors of ∠SRP and ∠PQR respectively.)

∠QPR = 2∠TRS - 2∠TQR

½ ∠QPR = ∠TRS - ∠TQR --------- (ii)

From (i) and (ii), we get

∠QTR = ½ ∠QPR

Hence, proved