In Fig. 6.44, the side $ \mathrm{QR} $ of $ \triangle \mathrm{PQR} $ is produced to a point $ \mathrm{S} $. If the bisectors of $ \angle \mathrm{PQR} $ and $ \angle $ PRS meet at point $ T $, then prove that $ \angle \mathrm{QTR}=\frac{1}{2} \angle \mathrm{QPR} $ "
Given:
The side $QR$ of $\triangle PQR$ is produced to a point $S$.
The bisectors of $\angle PQR$ and $\angle PRS$ meet at a point $T$.
To do:
We have to prove that $\angle QTR=\frac{1}{2}\angle QPR$.
Solution:
Let us consider the $\angle PQR$
We know that,
The sum of the interior angles is equal to the exterior angle.
Here, $\angle PRS$ is the exterior angle and
$\angle QPR$ and $\angle PQR$ are interior angles.
Therefore,
$\angle PRS=\angle QPR+\angle PQR$
This implies,
$\angle PRS-\angle PQR=\angle QPR$
Now, let us consider $\triangle QRT$
In a similar way we get,
$\angle TRS= \angle TQR+\angle QTR$
This implies,
$\angle QTR=\angle TRS-\angle TQR$......(a)
Since, $QT$ and $RT$ bisect $\angle PQR$ and $\angle PRS$ respectively. we get,
$\angle PRS=2\angle TRS$ and $\angle PQR= 2\angle TQR$
Therefore,
$\angle QTR=\frac{1}{2}\angle PRS-\frac{1}{2}PQR$
This implies,
$\angle QTR=\frac{1}{2}(\angle PRS-\angle PQR)$
from equation (a) we know that $\angle PRS-\angle PQR=\angle QPR$
Therefore,
$\angle QTR=\frac{1}{2}\angle QPR$
Hence proved.
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