jmc

Let us just drop the perpendicular from $B$ to $AC$ and label the point of intersection $O$. We will use this point later in the problem. As we can see, $M$ is the midpoint of $BC$ and $N$ is the midpoint of $HM$ $AHC$ is a $45-45-90$ triangle, so $\angle HAB=15^{\circ }$. $AHD$ is $30-60-90$ triangle. $AH$ and $PN$ are parallel lines so $PND$ is $30-60-90$ triangle also. Then if we use those informations we get $AD=2HD$ and $PD=2ND$ and $AP=AD-PD=2HD-2ND=2HN$ or $AP=2HN=HM$ Now we know that $HM=AP$, we can find for $HM$ which is simpler to find. We can use point $B$ to split it up as $HM=HB+BM$, We can chase those lengths and we would get $AB=10$, so $OB=5$, so $BC=5\sqrt{2}$, so $BM=\frac{1}{2}\cdot BC=\frac{5\sqrt{2}}{2}$ We can also use Law of Sines: $$\frac{BC}{AB}=\frac{\sin \angle A}{\sin \angle C}$$$$\frac{BC}{10}=\frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}}⟹BC=5\sqrt{2}$$ Then using right triangle $AHB$, we have $HB=10\sin 15^{\circ }$ So $HB=10\sin 15^{\circ }=\frac{5(\sqrt{6}-\sqrt{2})}{2}$. And we know that $AP=HM=HB+BM=\frac{5(\sqrt{6}-\sqrt{2})}{2}+\frac{5\sqrt{2}}{2}=\frac{5\sqrt{6}}{2}$. Finally if we calculate $(AP{)}^2$. $(AP{)}^2=\frac{150}{4}=\frac{75}{2}$. So our final answer is $75+2=77$. $m+n=\boxed{77}$.