Here are very natural approaches, not requiring hard theory: Solution 1 We want $\angle FMB = \angle MEB$. Let $T$ be the reflection of $C$ with respect to $M$ (so $ATBC$ is a parallelogram) - then the cyclic $AEDC$ gives $\angle ADE = \angle ACE = \angle ETB$, so $TBDE$ is cyclic. The parallelogram $ATBC$ and the midpoint $MF$ in triangle $TDC$ now yield \[\angle MEB = \angle MED + \angle DEB = \angle BAC + \angle DTB $$ $$ = \angle BAC + \angle MTB - \angle MTD = \angle BAC + \angle ACM - \angle CMF = \angle FMB. \] Solution 2 We want $\angle FMB = \angle MEB$. Clearly $\tan \angle FMB = \frac{FD}{DM} = \frac{CD}{2DM}$. To obtain info for $\tan \angle MEB$, put it in a right triangle - so let $BK\perp MC$ ($K\in MC$), hence $\tan \angle MEB = \frac{BK}{KE}$. The quadrilateral $AEBK$ is a parallelogram and so $BK = AE = AC\sin \varphi$ where $\angle ACM = \varphi$; also $KE = 2KM = 2\cos(\alpha + \varphi)MB = AB\cos(\alpha+\varphi) = \frac{AB \cdot DM}{CM}$. Hence $\frac{BK}{KE} = \frac{AC\cdot CM\cdot \sin \varphi}{2\cdot AM \cdot DM}$. It remains to justify $CD = \frac{AC \cdot CM \cdot \sin \varphi}{AM}$, but this follows from $CD = AC\sin\alpha$ and the Sine Law in triangle $ACM$.

Let $N$ be the midpoint of $AE$. The medians $MN$ and $MF$ are corresponding in the similar $\triangle AME$ and $\triangle CMD$, so $\angle FMD= \angle NME= \angle MEB$, done (we used that $MN$ is midline in $\triangle ABE$).

Let $X$ be reflection of $E$ over $M$, then $AXBE$ is parallelogram. Notice if $\triangle XEA \sim \triangle MDF$, then we have: $$ \angle AXE = \angle XEB = \angle FMD$$which solves the problem. So it is sufficient to show that: $$ \frac{AE}{EX} = \frac{DF}{MD} \implies AE \cdot MD =EX \cdot DF =ME \cdot CD \implies \frac{AE}{ME} = \frac{DM}{CD}$$The last line obviously holds as triangles $\triangle MEA \sim \triangle MDC$, hence we are done.

oops Make the problem $A$ indexed and WLOG $AB<AC$ instead (this doesn't actually end up mattering). Clearly $ACDE$ is cyclic. Place the problem in the coordinate plane with $A=(0,1),B=(b,0),C=(c,0)$, so $D=(0,0),M=(\tfrac{b+c}{2},0),F=(0,\tfrac{1}{2})$. We could just coordbash the rest of the problem, but we will not. In fact, we don't even need to compute $E$. Instead, define $f(\bullet)=\mathrm{Pow}_{(EMB)}(\bullet)-\mathrm{Pow}_{(F)}(\bullet)$. Clearly $f(D)=b\cdot \tfrac{b+c}{2}-\tfrac{1}{4}$. Now compute \begin{align*} f(A)&=(AM)(AE)-\frac{1}{4}\\ &=(AM)(AM+ME)-\frac{1}{4}\\ &=AM^2+MC\cdot MD-\frac{1}{4}\\ &=1+\left(\frac{b+c}{2}\right)^2+\frac{c-b}{2}\cdot\frac{b+c}{2}-\frac{1}{4}\\ &=\frac{bc+c^2}{2}+\frac{3}{4}. \end{align*}By linearity of power, it follows that $\mathrm{Pow}_{(EMB)}(F)=\tfrac{(b+c)^2}{4}+\tfrac{1}{4}=FM^2$, proving the desired tangency. $\blacksquare$

Assassino9931 wrote: Here are very natural approaches, not requiring hard theory: Solution 1 We want $\angle FMB = \angle MEB$. Let $T$ be the reflection of $C$ with respect to $M$ (so $ATBC$ is a parallelogram) - then the cyclic $AEDC$ gives $\angle ADE = \angle ACE = \angle ETB$, so $TBDE$ is cyclic. The parallelogram $ATBC$ and the midpoint $MF$ in triangle $TDC$ now yield \[\angle MEB = \angle MED + \angle DEB = \angle BAC + \angle DTB $$ $$ = \angle BAC + \angle MTB - \angle MTD = \angle BAC + \angle ACM - \angle CMF = \angle FMB. \] Another way to carry this angle chase out is to go from $\angle BAC+\angle DTB$ to $\angle DBT+\angle DTB=\angle MDT$, and then the conclusion follows from $\overline{MF} \parallel \overline{TD}$ since $\overline{MF}$ is the $C$-midline. Edit: ok this is more or less equivalent (still seems more morally correct?)