Given $\Delta ABC$ with its altitudes $AD,BE,CF$ and orthocenter $H$. Let $M,N,P$ the midpoint of $BC,AH,EF$. Let $G$ be the foot of perpendicular from $A$ to $EF$. Let $M'$ be the image of $M$ by reflection wrt $D$-midline of $\Delta DEF$. Prove $\angle AGN= \angle PGM'$
I thing angle chasing may solve the problem. But I kept chasing arround the conclusion. Also, the reflection wrt to the midline hypothesis is hard to approach. The only idea I have now is to prove $(NM'G)$ tangent to $EF$, but this is hard too.
Proof
It's well-known that $D,M,E,N,F~$lie on a circle, which is the nine-point circle of $\triangle ABC,$ and $A,E,H,F~$lie on another circle, whose center is $N$.
We may readily obtain that $M,P,M',N$ are collinear. From $D$ draw a line parallel to $EF$, which intersects $NM,AG$ at $Q,R$ respectively. Thus, $\triangle M'PG \cong \triangle MQR$. Let $AG$ intersect $BC$ at $T$, and $AD$ intersect $EF$ at $S.$ Obviously, $\triangle ASG \sim \triangle DTR.$
Notice that $\triangle FSN \sim \triangle DFN$. Thus$$\frac{AN}{NS}=\frac{NF}{NS}=\frac{ND}{NF}= \frac{ND}{AN}=\frac{DM}{MT}.$$
Hence, $\triangle ANG \sim \triangle DMR,$ which implies that $\angle PGM'=\angle DRM=\angle AGN.$