Question

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

M 

Answer 1

Step 1: Solve for the value of the $\angle QPR$and $\angle QTR.$

From$△PQR,$$\angle PRS$is an exterior angle.

$\angle QPR$and $\angle PQR$are interior angles.

$\angle PRS=\angle QPR+\angle PQR$(Exterior angle property)

$\angle PRS-\angle PQR=\angle QPR...\left(i\right)$

In $△QRT,$ $\angle TRS=\angle TQR+\angle QTR$(Since exterior angle are equal)

$\angle TRS-\angle TQR=\angle QTR$

Step 2: To prove that $QTR=\frac{1}{2}QPR.$

We know that $QT$and $RT$bisect $\angle PQR$ and $\angle PRS$ respectively.

So, $\angle PRS=2\angle TRS$ and $\angle PQR=2\angle TQR$

$$\begin{gathered} \angle QTR=\frac{1}{2}\angle PRS-\frac{1}{2}\angle PQR \\ \angle QTR=\frac{1}{2}\left(\angle PRS-\angle PQR\right) \end{gathered}$$

From equation $\left(i\right)$

we know that $\angle PRS-\angle PQR=\angle QPR$

$\angle QTR=\frac{1}{2}\left(\angle QPR\right)$

Hence proved that $\angle QTR=\frac{1}{2}\left(\angle QPR\right).$