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 the  side QR of triangle PQR is produced to a point S. If the bisectors of angle PQR and angle PRS meet at a point T, then prove that Angle QTR = 1/2 angle QPR.

I have an exam tomorrow!
I need the answer quick!!!!!

 

 

 

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 = ∠QPR

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This is the solution

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If the side QR of triangle PQR is produced to a point S. If the bisectors of angle PQR and angle PRS meet at the point T and angle QTR=21 degrees, then find the measure of angle QPR.
Please help me out in this question urgently....
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Good luck.

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Answer of questions

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Let angle TRS = a Angle PRQ = 180 – 2a Let angle TQR = b Therefore, angle PQT = b In triangle QPR Angle QPR = 2a – 2b = 2 (a – b) Similarly, in triangle QTR Angle QTR = a – b Therefore, angle QTR = ½ angle QPR Hence prove.
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