My solution: Let $O$ be the circumcenter of $\triangle ACE$. Easy to get $B, D, E$ be the symmetry of $O$ in $AC, CE, EA$ $\Rightarrow ABCO$ and $OCDE$ are the parallelograms $\Rightarrow AB\overset{\parallel }{=}OC\overset{\parallel }{=}ED$ $\Rightarrow ABED$ is the parallelogram $\Rightarrow BE$ passing through midpoint of $AD$ $(*)$ Similarly $CE$ passing through midpoint of $AD$ $(**)$. Since $(*), (**)$ deduced $AD$, $CF$ and $EB$ are concurrent(at midpoint of $AD$). DONE

Hi How did you prove that $B, D, E$ are the symmetry of $O$ in $AC, CE, EA.$

This can also be solved easily by Ceva's Theorem in trigonometric form.

Here's a different Solution:- 2015 PAMO P2 wrote: A convex hexagon $ABCDEF$ is such that $$AB=BC \quad CD=DE \quad EF=FA$$and $$\angle ABC=2\angle AEC \quad \angle CDE=2\angle CAE \quad \angle EFA=2\angle ACE$$Show that $AD$, $CF$ and $EB$ are concurrent. Solution:- Brianchon's Theorem reflects in our mind when we see that we have to prove $AD,BE,CF$ are concurrent, so before applying it we must prove that $ABCDEF$ has an inscribed circle. So let's prove it! Introduce the Circumcenter of $\triangle ACE$, let it be $O$. Let the perpendiculars from $O$ to $AB,BC,CD,DE,EF,FA$ be $l_1,l_2,l_3,l_4,l_5,l_6$ respectively. So it suffices to prove that $l_1=l_2=l_3=l_4=l_5=l_6$. Note that by the angle and length conditions we get that $\angle ABC=\angle AOC$ also $AB=BC$ and $AO=OC$. So, $AO=OC=AB=BC$.Similarly you get that $AO=OE=AF=FE$ and $CO=OE=ED=DC$. Hence, $AB=BC=CD=DE=EF=FA$. Now it's trivial to see that $l_1=l_2=l_3=l_4=l_5=l_6$. So, $ABCDEF$ must have an inscribed circle. So, by Brianchon we conclude.