In the given figure,in $\Delta PQR$,$\angle Q > \angle R$. $PM \perp QR$ and $PA$ is the bisector of $\angle QPR$. Prove that $\angle APM = \frac{1}{2}(\angle Q - \angle R)$.

(N/A) In $\Delta PQR$,let $\angle QPR = 2\alpha$. Since $PA$ is the angle bisector,$\angle QPA = \angle RPA = \alpha$.
In $\Delta PQR$,$\angle Q + \angle R + \angle QPR = 180^{\circ}$,so $\angle Q + \angle R + 2\alpha = 180^{\circ}$,which implies $\alpha = 90^{\circ} - \frac{1}{2}(\angle Q + \angle R)$.
In $\Delta PMQ$,$\angle PMQ = 90^{\circ}$,so $\angle QPM = 90^{\circ} - \angle Q$.
Now,$\angle APM = \angle QPA - \angle QPM$.
Substituting the values: $\angle APM = \alpha - (90^{\circ} - \angle Q)$.
$\angle APM = [90^{\circ} - \frac{1}{2}(\angle Q + \angle R)] - 90^{\circ} + \angle Q$.
$\angle APM = \angle Q - \frac{1}{2}\angle Q - \frac{1}{2}\angle R$.
$\angle APM = \frac{1}{2}\angle Q - \frac{1}{2}\angle R = \frac{1}{2}(\angle Q - \angle R)$.
Hence,proved.