Midpoints and other constraints in triangle I'm trying to solve this problem:

I can intuitively observe that $\angle IFH = 2\angle NMH$, but don't know how to prove it.
I extended segment $NM$ until it intersects the (extended) segment $IF.$ Let's call the intersection point $P.$
If I could prove that newly created triangle FMP is isosceles $(FM = FP)$, then this would prove that $\angle IFH = 2\angle NMH. $
I even did a simulation in Fusion 360 (video here) and it's interesting to see how the constraints work together to keep $FM = FP.$
 A: Let $K$ be a point on $IH$ such that $FK\parallel MN.$

$$\frac{HM}{MF}=\frac{HN}{NK}\implies \frac{a+b}{b}=\frac{c}{d}\implies ad+bd=bc .$$
We can manipulate the above expression to show,$$\frac{a}{a+2b}=\frac{c-d}{c+d}\implies \frac{FI}{FH}=\frac{IK}{HK}$$
By the converse of Angle Bisector Theorem,
$$\angle KFI=\angle KFH=\alpha \implies  \angle IFH=2\alpha.$$
A: I was just going through this question and I could see a simple solution that I would like to share.

If $Q$ is the midpoint of $FH$, $MQ = \frac{GH}{2}$. Also by midpoint theorem, $NQ \parallel IF$ and $NQ = \frac{IF}{2} = \frac{GH}{2}$.
As $MQ = NQ, \angle IFH = \angle NQH = 2 \alpha$
A: Rotate $\triangle FHI$ by $180^\circ$ about the side-midpoint $N$. Let the images of $F,M,G$ be $F', M', G'$ respectively.

After the rotation, $FHF'I$ is a parallelogram and $FH\parallel IF'$.
Join $FG'$, the segment $FG'$ split $\angle IFH$ into
$$\angle IFH = \angle IFG' + \angle G'FH.$$
Since lengths $FM=MG = M'G'$, $FMM'G$ is also a parallelogram and $FG' \parallel MN$. Then
$$\begin{align*}
\angle G'FH &= \angle NMH &&(\text{corresponding angles, }FG'\parallel MN)\\
 &= \alpha ^\circ
\end{align*}$$
Since lengths $FI=GH=IG'$, $\triangle FG'I$ is isosceles, and its base angles are equal:
$$\begin{align*}
\angle IFG' &= \angle IG'F &&(\text{base angles})\\
&= \angle G'FH &&(\text{alternate angles, }FH\parallel IF')\\
&= \alpha^\circ
\end{align*}$$
Together,
$$\angle IFH = \angle IFG' + \angle G'FH = 2\alpha^\circ$$
