Question

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

Answer 1

To Prove : $\angle QTR=\frac{1}{2}\angle QPR$.

Proof : $QT$ bisects $\angle PQR$ and $TR$ bisects $\angle PRS$

In $\Delta PQR$, $\angle RPQ+\angle PQR=\angle PRS$ (an exterior angle of a triangle is equal to the $\left(1\right)$ sum of the opposite interior angles)

also,

In $\Delta QRT$, $\angle RQT+\angle QTR=\angle TRS$ (an exterior angle of a triangle is equal $(20$ to the sum of the opposite interior angle)

From $\left(1\right)$ we have $\angle QPR=\angle PRS-\angle PQR=2\angle TRS-2\angle RQT=2(\angle TRS-\angle RQT)⟶\left(3\right)$

From $\left(2\right)$ we have $\angle QTR=\angle TRS-\angle RQT⟶\left(4\right)$

Comparing $\left(3\right)$ and $\left(4\right)$ we have

$\angle QPR=2\angle QTR$

$\Rightarrow \angle QTR=\frac{1}{2}\angle QPR$ Hence proved.