Draw triangle $ABC$ $(AB<AC)$, angle bisector $AD$ $(D\in BC)$ and midpoint $M$ of $BC$. From $C$ draw a perpendicular point to $AD$ at $H$; $HM$ intersect $AC$ at $N$; $CH$ intersects $AM$ at $E$; $ED$ intersects $AB$ at $F$.
a) Proof $N$ is the midpoint of $AC$.
b) Proof $DFA$ is an isosceles triangle (help me format the triangle).
Image: 
What have I do so far:
For a) I have to proof $MH//AB$ but I stucked
b) I have to proof $DE//AC$, yeah, but I think the proof would be same as a)
(a) Since angle $AHC$ is a right-angle, $H$ is on the circle with $AC$ as diameter.
Let $N'$ be the midpoint of $AC$. Then angles $AHN'$ and $HAN'$ are equal (isosceles triangle) and angles $HAB$ and $HAN'$ are equal (angle bisector) and therefore angles $AHN'$ and $HAB$ are equal.
Then $AB$ and $HN'$ are parallel. Therefore $HN'$ passes through $M$ and so $AB$ and $HM$ are parallel. (Note that we have now proved that $N=N'$.)
(b) There are various ways to complete the proof and you might like to have to have a go at this yourself.
One neat way is to note that, by Ceva's Theorem in triangle $ACH$, $AD.HE=DH.EC$ and so triangles $HDE$ and $HAC$ are similar.