Question

 the side QR of Δ PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, and then prove that ∠ QTR = ½ ∠ QPR.

Answer 1

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

HENCE PROVED

Answer 2

Hello mate ☺

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Solution:

∠PQT=∠TQR               (Given)

∠PRT=∠TRS               (Given)

To Prove:  ∠QTR=1/2(∠QPR)

∠PRS=∠QPR+∠PQR

(If a side of a triangle is produced, then the exterior angle is equal to the sum of two interior opposite angles.)

⇒∠QPR=∠PRS−∠PQR

⇒∠QPR=2∠TRS−2∠TQR

⇒∠QPR=2(∠TRS−∠TQR)

=2(∠TQR+∠QTR−∠TQR)                          (∠TRS=∠TQR+∠QTR)

(If a side of a triangle is produced, then the exterior angle is equal to the sum of two interior opposite angles.)

⇒∠QPR=2(∠QTR)

⇒∠QTR=1/2(∠QPR)

Hence Proved

I hope, this will help you.☺

Thank you______❤

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