Question

In Fig, the side $QR$ of $△PQR$ is produced to a point $S$.
If the bisectors of $\angle PQR$ and $\angle PRS$ meet at point $T$, then
prove that $\angle QTR$ = $\frac{1}{2}$ $\angle QPR$ .

Answer 1

In $△PQR$,

$\angle PQR+\angle QPR=\angle PRS$ [Exterior angle property]

$\Rightarrow 2\angle TQR+\angle QPR=2\angle TRS$ [As $TQ$ and $TR$ are bisectors of $\angle PQR$ and $\angle PRS$ respectively]

$\Rightarrow \angle QPR=2(\angle TRS-\angle TQR)$ ...(1)

Now, in $△TQR$,

$\angle TQR+\angle QTR=\angle TRS$ [Exterior angle property]

$\Rightarrow \angle QTR=\angle TRS-\angle TQR$...(2)

From (1) and (2)

$\Rightarrow \angle QTR=\frac{\angle QPR}{2}$

Hence, Proved