See here:http://www.artofproblemsolving.com/community/c6h1115995_strange_condition.

Let S be point on BC such that BS = BA and CS = CD. Note that EB and EC are perpendicular bisectors of AS and DS so E is center of circle ADS. ∠AED = 2∠ASD = 2(∠ABS/2 + ∠DCS/2) = ∠ABS + ∠DCS = 180 - ∠AFD so AFDE is cyclic.

Easy Constructive Geo Let $E$ be the intersection of the angle bisectors $A'$ be on $BC$ Such that $AB=A'B$ and $CD=A'C$ Join $A,A'$, $A'E$, $A',D$, $E,D$ and $A,E$ By angle chasing, $AA'$ is perpendicular to $BE$ and $CE$ is perpendicular to $A'D$ Now just prove $E$ is the circumcenter of $\triangle AA'D$ [Show $\triangle A'BE$ and $\triangle ABE$ are congruent. And $\triangle A'CE$ is congruent to $\triangle CDE$ ] Let $\angle A'BE=x$, $\angle A'CE=y$ and $\angle EA'A=z$ this leads to $\angle AFD=180-2x-2y$ and $\angle AED=2x+2y$ [Just Find all angles in terms of $x,y,z$]