Question

In Fig.  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 ∠QTR = \(\frac{1}{2}\)∠QPR.

Answer 1

Data: Arm QR of ∆PQR is produced upto S. Angular bisectors of ∠PQR and ∠PRS meet at T. 

To Prove: ∠QTR = \(\frac{1}{2}\)∠QPR

Proof: 

Let ∠PQR = 50°, and ∠PRS = 120°, 

∠PQT = ∠TQR = 25° 

∠PRT = ∠TRS = 60° 

QR arm of ∆PQR is produced upto S. 

∴Exterior angle ∠PRS = ∠PQR + ∠QPR 

120 = 50 + ∠QPR 

∴ ∠QPR =120 – 50 

∠QPR = 70°

∴ ∠PRQ = 60° 

Now, in ∆TRQ, 

∠TQR + ∠TRQ +∠QTR = 180° 

25 + 120 + ∠QTR = 180° 

145 + ∠QTR = 180° 

∠QTR = 180 – 145 

∴ ∠QTR = 35° 

Now, ∠QTR = 35° 

∠QPR = 70° 

∠QTR = \(\frac{70}{2}\)

∴ ∠QTR = \(\frac{10}{2}\)x ∠ QPR.