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# 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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