Question

In the figure below, $\angle Q>\angle R$ and $M$ is a point on $QR$ such that $PM$ is the bisector of $\angle QPR$ If the perpendicular from $P$ on $QR$ meets $QR$ at $N$ then prove that $\angle MPN=\frac{1}{2}(\angle Q-\angle R)$

Answer 1

$\because \angle QPM=\angle RPM$

In $\Delta PNQ$

$\angle QPN={90}^{\circ }-\angle Q$

In $\Delta PNR$

$\angle RPN={90}^{\circ }-\angle R$

$\angle QPM=\angle RPM$

$\Rightarrow \angle QPN+\angle MPN=\angle RPN-\angle MPN$

$\Rightarrow 2\angle MPN=\angle RPN-\angle QPN$

$\Rightarrow 2\angle MPN=({90}^{\circ }-\angle R)-({90}^{\circ }-\angle )$

$\Rightarrow \angle MPN=\frac{1}{2}(\angle Q-\angle R)$ (proved)