Neat! Claim 1: $XAFE$ is cyclic Proof: Indeed, note that simce $AX=AM$, $\angle FBX=\angle ABX=\angle ACX=\angle ECX$ which gives us $\angle AEX=\angle 2 \cdot \angle ECX=2 \cdot \angle FBX=\angle AFX$, proving the claim. Let $XE \cap (ABC)=Y \neq X$. Observe that since $\angle YXC=\angle MCA$, we get $YC=AM$, implying that $MYCA$ is an isosceles trapezium. $\implies \angle AFE=\angle AXE=\angle AXY=\angle ACY=\angle MAC=\angle MBC=90-\frac{A}{2}$ $\implies \angle AEF=90-\frac{A}{2}=\angle AFE$ $\implies AE=AF$, as required.$\blacksquare$

Firstly, $\triangle XBF$ is similar to $\triangle EXC$ by spiral similarity. Therefore $AXFE$ is cyclic. We now finish by angle chase: $\angle AEF=\angle AEX+\angle XEF=\angle AFX +\angle FAX \\ =2\angle ABX+ \angle BAX = 2\angle ACX + \angle XCB = \angle MCB $. Therefore: $\angle AFE =180- \angle FAE - \angle FEA =180- \angle BMC - \angle MCB = \angle MBC = \angle MCB $. Hence $\angle AEF = \angle AFE $ and the assertion follows.