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  • In Fig. 6.44, 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 = 1 / 2 ∠ QPR.

6. In Fig. 6.44, the side QR of \Delta 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 = \frac{1}{2}   \angleQPR.

                

Answers (1)

best_answer

We have, 
\Delta PQR is produced to a point S and  bisectors of \angle PQR and \angle PRS meet at point T,
By exterior angle sum property,
\angle PRS = \angleP + \anglePQR
Now, \frac{1}{2}\angle PRS =\frac{1}{2}\angle P + \frac{1}{2} \angle PQR
\Rightarrow \angle TRS = \frac{1}{2}\angle P + \angle TQR................(i)

Since QT and QR are the bisectors of  \angle PQR and \angle PRS respectively.

Now, in \DeltaQRT,
\angle TRS = \angle T + \angle TQR..............(ii)

From eq (i) and eq (ii),  we get

\frac{1}{2}\angle P= \angle T
\Rightarrow \frac{1}{2}\angle QPR= \angle QTR
Hence proved

Posted by

manish painkra

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