Since $AEID$ is parallelogram $AI$ bisects $DE$ and since $A, I , F$ are collinear we have that $FI$ bisects $DE$. In $\triangle FDE: BE,DC,FN$ are concurrent at $I$ ($N$ is midpoint of $DE$) thus by Cevas theorem $BC || DE$ (since $DN = EN$). Now we have that $AI$ also bisects $BC$ and since $AI$ is angle bisector of $\angle BAC$ we have $AB = AC.$

Since $AE || CD $ and $AD || BE $, so, $\frac {FC}{CE} $ = $\frac {FI}{IA} $ = $\frac {FB}{BD} $. Thus, $BC || DE$. Simple angle chasing proves that $ADBCE $ is cyclic and hence the result follows.

We use barycentric coordinates. As $D$ lies in $\angle ACB$ angle bisector, we have that $D=(a:b:t)$ for some $t$, therefore $AD$ is given by the equation $y:z=b:t$. We know that $BI$ is given by $x:z=a:c$. Since $BI\|AD$, we have that $BI\cap AD$, is the point at infinity, hence the solutions of the system $y:z=b:t ; x:z=a:c$ must be with $x+y+z=0$, then, by easy calculations, we find that $t=\frac{-cb}{a+c} \Rightarrow D=(a:b:\frac{-cb}{a+c})$ and similarly $E=(a:\frac{-cb}{a+b}:c)$. Taking $BD \cap CE$, we find that $F=(a(a+b)(a+c):-cb(a+c):-cb(a+b))$. We know that $$F\in AI \Leftrightarrow (-cb)(a+c):(-cb)(a+b)=b:c$$$$\Leftrightarrow ac+c^{2}=ab+b^{2}$$$$\Leftrightarrow a(c-b)=-(b+c)(c-b)$$$$\Leftrightarrow b=c $$(Since if $b\neq c \Rightarrow 0<a=-(b+c)<0$, contradiction), as desired.

Clearly $AI$ bisects $DE$, and $DC$,$BE$,$FI$ are concurrent,so $BC$ is parallel to $DE$, so $BC$ also is bisected by $AI$, but $AI$ is angle bisector of angle $BAC$,so we are done.

Just simple angle chasing: $\angle DAB$ = $\angle ABI$ = $\angle CAE$ = $\angle ACD$. And we know $BI$ and $CI$ are angle bisectors $\Rightarrow$ $\angle ABC$ = $\angle ACB$ $\Rightarrow$ $AB=AC$. Hence Proved. P.S: I wonder why the unnecessary observation $DE || BC$ is required, as seen in all the other solutions.

mather007 wrote: Just simple angle chasing: $\angle DAB$ = $\angle ABI$ = $\angle CAE$ = $\angle ACD$. And we know $BI$ and $CI$ are angle bisectors $\Rightarrow$ $\angle ABC$ = $\angle ACB$ $\Rightarrow$ $AB=AC$. Hence Proved. P.S: I wonder why the unnecessary observation $DE || BC$ is required, as seen in all the other solutions. You assumed that $E$&$D$ lies on the circumcircle which you should prove. In fact you used the fact that $AB=AC$ to prove it!!

From the parallel conditions we have that $ADIE$ is a paralelogram. Since F lies in the perpendicular bisector of $DE$, $FD=FE$. Now from Thales Theorem, $\frac{AD}{IB}=\frac{AF}{IF}$ and $\frac{AE}{CI}=\frac{AF}{IF}$ so $\frac{AD}{IB}=\frac{AE}{IC}$ and from angle-chasing, $$\angle DAE=\angle DIE=\angle BIC$$so $\triangle BIC~\triangle DAE$. So a homothety from F sents $\triangle BIC$ to the isosceles triangle $\triangle DAE$. So we get that $\triangle BIC$ is an isosceles triangle therefore so is $\triangle BAC$