So easy, just take a point $P$ on $AD$, such that $AB=AP, CD=PD$.

WLoG $AB<CD$.Let $BF$ meet $CD$ at $Q$.Since $FE\parallel CD$,$\frac{AF}{FD}=\frac{AE}{EC}$.Again since $AB\parallel CD$,$\bigtriangleup AEB$~$\bigtriangleup DEC$,hence $\frac{AE}{EC}=\frac{AB}{CD}$.The above two equations imply $FD=CD$.We have $AB+CD=AD=AF+FD$,so,$AF=AB$.Again $\bigtriangleup ABF$~$\bigtriangleup QFD$,so,$\frac{AF}{FD}=$$\frac{AB}{QD}$ implying $FD=QD$.So,$D$ is the circumcenter of $\bigtriangleup QFC$ and it lies on $CQ$,so $\sphericalangle QFC=$$90^{\circ}$. Q.E.D

$\color{blue} \boxed{\textbf{SOLUTION}}$ Let, $F' \in AD$ s.t. $AB=AF',CD=BF'$ $BF' \cap CD \equiv B', CF' \cap AB \equiv C'$ Now, $\angle AF'B=\angle ABF'=\alpha \implies \angle F'AB=180°-2\alpha \implies F'DC=2\alpha$ $\implies \angle DF'C=\angle DCF'=90°-\alpha$ So, $\angle BF'C=90°$ $\angle C'F'B=\angle C'F'A + \angle AF'B= 90°-\alpha+\alpha=90°$ And as, $AF'=AB \implies AC'=AF'=AB$ Similarly, $B'D=CD=F'D$ $\triangle ABF'$~$\triangle B'DF'$ $\implies \frac{DF'}{F'A} = \frac{B'D}{AB}=\frac{CD}{AB}$ $\triangle CED$~$\triangle AEB$ $\implies \frac{CD}{AB}=\frac{DE}{EB}$ So, $\frac{DF'}{F'A}=\frac{DE}{EB} \implies EF' \parallel AB \parallel CD \implies F' \equiv F$ And, $\angle BF'C=90°=\angle BFC \blacksquare$