Question

6. In fig.6.44, the side QR of triangle PQR is produced to a point S. If the bisectors of < PQR and < PRS meet at point T , then prove that < QTR = 1/2 < QPR.​

Answer 1

Given

• Bisectors of < PQR and < PRS meet at point T.

Prove that

• < QTR = 1/2 < QPR.

To Prove,We have

QT is the bisector of <PQR

• 1/2 PQR=PQT=TQR

TR is the bisector of PRS

• 1/2PRS= PRT=SRT

Now In Trinagle PQR

PRQ is a Exteriors angel

PRS = QPR + POR [Exterior angle property]--(1)

Now in triangle QTR

TRS is the external angle

TRS = TQR + QTR [Exterior angle property]--(2)

Putting TRS = 1/2PRS & TQR = 1/2PQR

• 1/2 PRS = 1/2PQR + QTR
• 1/2 PRS = 1/2PQR + QTR

Putting PRS = QPR + PQR from ( 1 )

$\sf \: \dfrac{1}{2} (QPR+PQR)= \dfrac{1}{2} PQR+QTR$

$\sf \: \dfrac{1}{2}QPR+\dfrac{1}{2}PQR=\dfrac{1}{2}PQR+QTR$

$\sf \: \dfrac{1}{2}QPR+\dfrac{1}{2}PQR-\dfrac{1}{2}PQR=QTR$

$\sf \: \dfrac{1}{2}QPR=QTR$

$\sf QTR = \: \dfrac{1}{2}QPR$

Proved