Question

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

Answer 1

Given

$TQ$ is the bisector of $\angle PQR$

So, $\angle PQT=\angle TQR=\frac{1}{2}\angle PQR$

Also,

$TR$ is the bisector of $\angle PRS$

So, $\angle PRT=\angle TRS=\frac{1}{2}\angle PRS$

In $PQR$

$\angle PRS$ is the external angle

$\angle PRS=\angle QPR+\angle PQR..............\left(1\right)$ [External angle is sum of two interior opposite angles]

In $TQR$

$\angle TRS$ is the external angle

$\angle TRS=\angle TQR+\angle QTR..............\left(2\right)$ [External angle is sum of two interior opposite angles]

Putting $\angle TRS=\frac{1}{2}\angle PRS\&\angle TQR=\frac{1}{2}\angle PQR$

$\frac{1}{2}\angle PRS=\frac{1}{2}\angle PQR+\angle QTR$

Putting $\angle PRS=\angle QPR+\angle PQR$ from (1)

$\frac{1}{2}(\angle QPR+\angle PQR)=\frac{1}{2}\angle PQR+\angle QTR$

$\frac{1}{2}\angle QPR+\frac{1}{2}\angle PQR=\frac{1}{2}\angle PQR+\angle QTR$

$\frac{1}{2}\angle QPR+\frac{1}{2}\angle PQR-\frac{1}{2}\angle PQR=\angle QTR$

$\angle QTR=\frac{1}{2}\angle QPR$

Hence proved.