[asy][asy] import olympiad; size(7cm); defaultpen(fontsize(10pt)); pair A=(0,0), B=dir(230), E=dir(310); pair X=extension(B,B+rotate(90)*(B), A,E), Y=extension(E,E+rotate(90)*(E), A,B); pair P=intersectionpoint(B--X, E--Y); pair F=(0,-1); pair D = extension(B,F,E,P), C=extension(E,F, B,P); pair Q=intersectionpoint(A--F, B--E); draw(A--B--P--E--A); draw(circumcircle(B,E,-B)); draw(C--D); draw(B--D); draw(C--E); draw(B--E); draw(A--P); label("$A$", A, (0,1)); label("$B$", B, (-1,-1)); label("$C$", C, (-1,-1)); label("$D$", D, (1,-1)); label("$E$", E, (1,-1)); label("$F$", F, (0.5,1)); label("$P$", P, (0,-1)); label("$Q$", Q, (1,1)); [/asy][/asy] Claim: $\overline{BE} \parallel \overline{CD}$ Proof: Construct circle $\omega$ with center $A$ and it passes through $B$. Since $AB = AE$, $\omega$ passes through $E$ as well. Since $m\angle{ABC} = m\angle{ADE} = 90^{\circ}$, $BC$ and $DE$ are tangent to $\omega$. Now, let $\overline{BC} \cap \overline{DE} = P$. This means that $BP = EP$. We also know that $BC = DE$. This means that $CP = DP$. Since $B,C,P$ are collinear and $E,D,P$ are collinear, we can see that $m\angle{CPD} = m\angle{BPE}$. Now, let $m\angle{CPD} = \theta$. Since we know that $\Delta{CPD}$ and $\Delta{BPE}$ are both isosceles and share $\angle{BPE}$, we can see that $m\angle{DCP} = m\angle{EBP} = m\angle{BEP} = m\angle{CDP} = 90^{\circ}-\frac{\theta}{2}$. Since $m\angle{EBP} = m\angle{CDP}$, we get that $\overline{BE} \parallel \overline{CD}$. This completes our proof that $\overline{BE} \parallel \overline{CD}$. Claim: $\overline{BD}$ bisects $\angle{EBC}$ Proof: Since $BC = CD$, we get that $\Delta{BCD}$ is isosceles which means that $m\angle{CBD} = m\angle{CDB}$. We also know that $m\angle{EBD} = m\angle{BCD} = m\angle{CBD}$. This means that $m\angle{CBD} = m\angle{EBD}$ which means that $\overline{BD}$ bisects $\angle{EBC}$. This completes our proof that $\overline{BD}$ bisects $\angle{EBC}$. Similar logic can be used to show that $\overline{CE}$ bisects $\angle{BED}$. Claim: $\overline{AF} \perp \overline{BE}$ Proof: Recall that $BC = DE$ and $\overline{BE} \parallel \overline{CD}$. This means that $BCDE$ is an isosceles trapezoid. This means that $m\angle{CBE} = m\angle{DEB}$. Recall that $\overline{BD}$ and $\overline{CE}$ are angle bisectors. This means that $m\angle{CBD} = m\angle{BDE} = \frac{1}{2}m\angle{CBE} = \frac{1}{2}m\angle{BED} = m\angle{DEC} = m\angle{BEC}$. This means that $m\angle{BEC} = m\angle{EBD}$ which means that $\Delta{BFE}$ is isosceles. This means that $BF = FE$. But, recall that $AB = AE$. This means that $ABFE$ is a kite. Since $ABFE$ is a kite, we know that its diagonals are perpendicular which means that $\overline{AF} \perp \overline{BE}$. Now, let $m\angle{CBD} = \alpha$. By some simple angle chasing, we get that $m\angle{ABE} = 90-2\alpha$. Since $m\angle{CBD} = m\angle{EBF} = \alpha$, we get that $m\angle{BAF} = 90^{\circ}-\alpha$. Now, since we know that $m\angle{AQB} = 90^{\circ}$ and $m\angle{ABQ} = 90^{\circ}-2\alpha$, we get that $m\angle{BAF} = 2\alpha$. But, we know that $m\angle{ABF} = 90^{\circ}-\alpha$. This means that $m\angle{ABF} = m\angle{AFB} = 90^{\circ}-\alpha$. This means that $\Delta{BAF}$ is isosceles which means that $AB = AF$. Q.E.D. P.S. Let me know if anyone has a better solution to proving $F$ lies on $\omega$