Let $FE$ intersect $BC$ at point $P$. [asy][asy] size(8cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = A+C-B; pair E = extension(A,C,B,incenter(A,B,C)); pair F = intersectionpoints(A--11*A-10*D,B--B+10*(E-B)*dir(90))[0]; draw(A--B--C--D--A--cycle); draw(A--C); draw(A--F); draw(B--F); draw(B--E); pair P = extension(F,E,B,C); pair M = (B+E)/2; draw(F--P); draw(C--P); draw(C--M); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$P$",P,dir(P)); dot("$M$",M,dir(M)); [/asy][/asy] First, note that $$\angle FAB=\angle ABC=180^{\circ}-2\angle ABF$$Thus, $AF=AB$. Next, we show that $BC=CP$, which would imply that $MC\parallel EP$. We have $$\frac{BC}{BA}=\frac{CE}{EA}=\frac{CP}{AF}=\frac{CP}{BA}$$from which the result follows.

Let $T\in BF, AT$ bisects $\angle BAF$. Then $AT\parallel BE\perp BF$. Hence $TM\parallel EF$. Let $TM$ meet $BC$ at $C'$. Then by Pappus on $\overline{B,T,F}$ and $\overline{\infty,\infty,\infty}$ we have $C'\in AE$.

Let $\ell$ be the line through $C$ parallel $BE$. $$(P_\infty,A,K=CM\cap AD,L=\ell\cap AD)=(B,E,M,P_\infty)=-1$$Thus, $AK=AL$. Also, $$(F,X=BE\cap AD,A,P_\infty)=(BF\cap AC,E,A,C)=-1,$$hence $AF=AX$. Now, $$\frac{AE}{AC}=\frac{AX}{AL}=\frac{AF}{AK},$$we are done.

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