Question

Question 6
In the given figure, the side QR of $\Delta PQR$ is produced to a point S. If the bisector of $\angle PQRand\angle PRS$ meet at point T, then prove that $\angle QTR=\frac{1}{2}\angle QPR.$

Answer 1

Considering $\Delta QTR,\angle TRS$ is an exterior angle.
$\therefore \angle QTR+\angle TQR=\angle TRS$ (exterior angle is equal to sum of opposite interior angles)
$\angle QTR=\angle TRS-\angle TQR$……………. (1)
$For\Delta PQR,\angle PRS$ is an external angle.
$\therefore \angle QPR+\angle PQR=\angle PRS$ (exterior angle is equal to sum of opposite interior angles)
$\angle QPR+2\angle TQR=2\angle TRS$ (As QT and RT are angle bisectors, $\angle PQR=2\angle TQRand\angle PRS=2\angle TRS)$
$\angle QPR=2(\angle TRS-\angle TQR)$
$\angle QPR=2\angle QTR$ [ By using equation (1)]
$\therefore \angle QTR=\frac{1}{2}\angle QPR$