Quadrilateral with two congruent legs of diagonals I've come across a geometry proof which seems like it should be easy, but I'm struggling with it:
Suppose you have a convex quadrilateral $ABCD$ whose diagonals intersect at $E$.  Given that angles $ABC$ and $BCD$ are congruent, and the two "legs" of the diagonals $EA$ and $ED$ are congruent, show that the sides $BC$ and $AD$ are parallel (that is, show that this quadrilateral is an isosceles trapezoid).
(If this cannot be proven, please explain why or give a counterexample.)
I know that we can use the isosceles triangle theorem on triangle $AED$ to show that angles $EAD$ and $EDA$ are congruent, and I know that the angles $BEA$ and $CED$ are congruent, as well as $BEC$ and $DEA$.  I'm trying to get a pair of congruent triangles that will show that the sides $AB$ and $CD$ are congruent, but I'm always missing one constraint. 
For example, if I try to show that triangles $ABC$ and $DCB$ are congruent, I have a side $BC$ congruent to itself, and two angles $ABC$ and $DCB$ congruent, but I can't show that the diagonals are congruent because I only know that $AE$ is congruent to $DE$, and I don't know that $EB$ is congruent to $EC$.
If I try to show that triangles $AEB$ and $DEC$ are congruent, I have side $AE$ congruent to side $DE$, and angles $AEB$ and $DEC$ congruent, but I can't show that any other pair of angles is congruent, since I only know angles $EAD$ and $EDA$ are congruent and angles $ABC$ and $DCB$ are congruent, but I can't show that angles $EBA$ and $ECD$ are congruent.
I have a strong intuition that this proposition is true, but I can't prove it.  I'm hoping someone here can either give a sound proof or give a counterexample that explains why this proposition is false.
 A: Can we work with the following fact?
Because $\Delta XBC$ is an Isosceles triangle, the only way $A'E_2$ and $E_2 D$ (or $A E_1$ and $E_1 D'$) can be of equal length, is that $A'$ and $A$ (or $D'$ and $D$) coincide.
Therefore, $\angle XAD$ must be equal to $\angle XDA$ for $AE$ to be equal to $DE$. And since $\angle XAD + \angle XDA = \angle XBC + \angle XCB$, it is evident that $AD$ and $BC$ are parallel.

A: $\def\C{{\cal C}}$
Let $\C_A$ and $\C_D$ be the circumscribed circles of $\triangle ABC$ and
$\triangle DBC$, respectively.
Assume these circles do not coincide: $\C_A\ne\C_D$.
Since the circles intersect at points $B$ and $C$ one of them is inside the other in the upper half-plane (the half-plane created by the line $BC$, where the points $A$ and $D$ are situated). Without loss of generality we may assume it is the circle $\C_D$. This means that $D$ is an inner point of the circle $\C_A$ (see figure).
Continue the lines $CD$ and $AD$ till intersection with the circle $\C_A$.
Let $F=(CD)\cap\C_A$, $G=(AD)\cap\C_A$. Observe that $G$ is an inner point of the arc $FC$.

In view of $\angle ABC=\angle BCF$ and $\angle BAC=\angle BFC$ we have:
$$
\triangle ABC\cong \triangle FCB\implies \angle BCA=\angle CBF.
$$
Hence:
$$
\angle DAE=\angle GAC=\frac12 \overset{\mmlToken{mo}{⏜}}{GC}
<\frac12 \overset{\mmlToken{mo}{⏜}}{FC}
=\frac12\overset{\mmlToken{mo}{⏜}}{AB}<\angle ADB=\angle ADE,
$$
which contradicts to the condition that $AED$ is isosceles triangle.
Thus, the assumption $\C_A\ne \C_D$  was false, and the quadrilateral $ABCD$ is in fact cyclic. The conclusion $AD\parallel BC$ follows immediately.
