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

(N/A) Given: In $\triangle PQR$,$\angle Q > \angle R$,$PA$ is the bisector of $\angle QPR$ and $PM \perp QR$.
To prove: $\angle APM = \frac{1}{2}(\angle Q - \angle R)$.
Proof:
$1$. Since $PA$ is the bisector of $\angle QPR$,we have $\angle QPA = \angle APR = \frac{1}{2} \angle QPR$.
$2$. In $\triangle PQM$,$\angle Q + \angle PMQ + \angle QPM = 180^{\circ}$. Since $\angle PMQ = 90^{\circ}$,$\angle Q + 90^{\circ} + \angle QPM = 180^{\circ}$,so $\angle Q = 90^{\circ} - \angle QPM$.
$3$. In $\triangle PRM$,$\angle R + \angle PMR + \angle RPM = 180^{\circ}$. Since $\angle PMR = 90^{\circ}$,$\angle R + 90^{\circ} + \angle RPM = 180^{\circ}$,so $\angle R = 90^{\circ} - \angle RPM$.
$4$. Subtracting the two equations: $\angle Q - \angle R = (90^{\circ} - \angle QPM) - (90^{\circ} - \angle RPM) = \angle RPM - \angle QPM$.
$5$. We can write $\angle RPM = \angle APM + \angle APR$ and $\angle QPM = \angle QPA - \angle APM$.
$6$. Substituting these: $\angle Q - \angle R = (\angle APM + \angle APR) - (\angle QPA - \angle APM)$.
$7$. Since $\angle QPA = \angle APR$,this simplifies to $\angle Q - \angle R = 2 \angle APM$.
$8$. Therefore,$\angle APM = \frac{1}{2}(\angle Q - \angle R)$. Hence proved.