Let $N$ be the midpoint of $AE$. We will show that $NM \perp BC$. Denote the midpoint of $BD$ as $S$. Notice that triangles $AME$ and $BMD$ are similar, also since $N$ is the midpoint of $AE$ and $S$ is the midpoint of $BD$, triangles $AMN$ and $BMS$ are similar. Thus, $\angle{NMB} = \angle{AMS}$. We also have $MS // DC$ and $DC \perp AD$ so $MS \perp AD$, so $\angle{NMB} = \angle{AMS} = 90^{\circ}$, or $NM \perp BC$, and we are done.

$ABDE$ cyclic $\implies$ $\angle FEM=\angle MDB$ and $\angle FAM=\angle MBD$ $FB=FC \implies \angle FME=90^o=\angle MDC$ $$\frac{MB}{MC}\cdot\frac{\sin\angle MDC}{\sin\angle MDB}=\frac{\sin\angle MDC}{\sin\angle MDB}=\frac{\sin\angle MCD}{\sin\angle MBD}$$ $$\frac{AF}{FE}\cdot\frac{\sin\angle FME}{\sin\angle AMF}=\frac{\sin\angle FEM}{\sin\angle FAM}=\frac{\sin\angle MDB}{\sin\angle MBD}=\frac{\sin\angle MDC}{\sin\angle MCD}$$ $$\frac{AF}{FE}\cdot\frac{\sin\angle FME}{\sin\angle AMF}=\frac{\sin\angle MDC}{\sin\angle MCD}=\frac{\sin\angle FME}{\sin\angle AMF}$$ $$\frac{AF}{FE}=1 \implies AF=FE \quad \blacksquare$$

$N$ be the foot of the perpendicular from $A$ to $BC$.$\angle ANC=\angle ADC=90^o$ So $ANDC$ cyclic. $M$ is on $AD$-radical axis of $(ANDC)$ and $(AEDB)$. So $MB\cdot ME=MC\cdot MN$ implies $ME=MN$. $FM$ is perpendicular bisector of $NE$. $FN=FE$. $ANE$ is right triangle, so $FN=FE=AF$ done.

Coordinate bash solution. Let $M=(0,0),B=(-1,0),C=(1,0),A=(a,b)$ then line $MA=\frac{b}{a}x$ and line $CD=-\frac{a}{b}(x-1)$.We know that $D$ is on $AM$.So, plugging same values for two equations will became $D=(\frac{a^2}{a^2+b^2},\frac{ab}{a^2+b^2})$. From, PoP $MA.MD=MB.ME$. $MA=\sqrt{a^2+b^2},MD=\frac{a}{\sqrt{a^2+b^2}},MB=1$, thus, $ME=a$ which tells us $E=(-a,0)$. So, equation of line $AE=\frac{b}{2a}x+\frac{b}{2}$. Also, $F$ is on $AE$ and $y$ axis. Let's say $F=(0,f)$ plugging values of $F$ to line equation of $AE$ shows $F=(0,\frac{b}{2})$ which is trivially says $F$ is midpoint of $AE$.

Median sine at $\Delta DBC$ and $\Delta FAM$ kill this problem.

mahmudlusenan wrote: Median sine at $\Delta DBC$ and $\Delta FAM$ kill this problem. I think median sine on triangles DBC and AEM

We note that <FBC=<FCB=α,<AEB=<ADB=β from cyclic and <CMD=θ.And we get from angle-chasing <BCD=90-θ,<CBD=θ-β=<EAD.We write median sine on triangle DBC because DM is median .And this median sine is same on triangle FAM.We get from this FM is median and AF=FE.We are done.

Attachments: