$ N: AE \cap CM$, $ G: CF \cap BA$. For Menelaus on CMB with trasversal NE we obtaid $ \overline{CN}=\overline{NM}$ and for Ceva on ACM and point F we obtain $ \overline{AG}=2 \overline{GM}$ and so $ \overline{GB}=\overline{GA}$. For Menelaus on AEB and trasversal GF we have $ \frac{\overline{AF}}{\overline{FE}}= \frac{3}{2}$ and so $ \frac{\overline{AF}}{\overline{AE}}= \frac{3}{5}$. Let $ CA=CB=x$ so $ \overline{AD}=\frac{2}{3} x$ and $ \overline{AE}= \frac{\sqrt{10}}{3} x$ so $ \frac{\overline{AE}}{\overline{AD}} = \frac{\sqrt{10}}{2}= \frac{\overline{AC}}{\overline{AF}}$ so AFD and ACE are similar and $ AE \perp DM$, then DCEF is cyclic with DC = CE so C stay on the angle bisector of $ \angle DFE$

See that AM^2-AD^2=ME^2-DE^2, therefore AE and DM are perpendicular, CDFE cyclic with CD=CE, so CF is angle bisector of DFE. Best regards, sunken rock

Let $C(0,0)$, $B(0,a)$ and $A(a,0)$. Easily, $D(\dfrac{a}{3},0)$, $E(0,\dfrac{a}{3})$, $M(\dfrac{a}{2},\dfrac{a}{2})$ Let $F(p,q)$. By solving a system, $\begin{vmatrix} p & q &1 \\ \dfrac{a}{2} & \dfrac{a}{2} & 1 \\ \dfrac{a}{3} & 0 & 1 \end{vmatrix}=0$ $\begin{vmatrix} p & q &1 \\ 0 & \dfrac{a}{3} & 1 \\ a & 0 & 1 \end{vmatrix}=0$ we easily get $p=\dfrac{2a}{5}$ and $q=\dfrac{a}{5}.$ $k_{EF}=-\dfrac{1}{3}$ $k_{CF}=\dfrac{1}{2}$ $k_{DF}=3$ $tg(\angle EFC)=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{1-\dfrac{1}{2}\dfrac{1}{3}}=1$ $tg(\angle CFD)=\dfrac{3-\dfrac{1}{2}}{1+3\cdot\dfrac{1}{2}}=1$ $\implies \angle EFC=\angle CFD$ and we are done.

Wait isnt this completely obvious by symmetry or am I messing up my diagram? Isosceles right triangle?

Oh wait nvm misread problem.

Because reviving the thread: Take $N$ symmetrical of $B$ about $C; D$ is the centroid of $\triangle ABN\implies N-D-M$ collinear, so $\angle ADM=\angle CDN$, but $\angle CDN=\angle BDC$ by symmetry; also by symmetry $\angle AEC=\angle CDB$, so $\angle ADM=\angle FEC$ and $CDFE$ is cyclic; with $CD=CE$ we are done. Best regards, sunken rock