Question

In Fig.6.44, the side QR of ∆PQR is produced to a point S.If the bisectors of angle PQR and angle PRS meet at point T,then prove that angle QTR =1/2 angle QPR.

Answer 1

here
∠QPT = ∠TQR ( LET THEM BE 'X')
∠QRT = ∠TRS ( LET IT BE 'Y')

THEN
IN TRIANGLE PQR
∠QPR + 2X + (180° -2Y) = 180° ∠QPR + 2X - 2Y = 0
∠QPR = 2Y - 2X   
∠QPR/2 = Y - X    -----(1)
IN TRIANGLE QRT
X + (180° - Y) + ∠QTR = 180°X - Y + ∠QTR = 0∠QTR = Y - X    --------(2)
USING 1 AND 2 WE GET∠QPR/2 = ∠QTR
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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