# 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.

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.

$$\def\C{{\cal C}}$$ 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 $$\angle ABC=\angle BCF$$ we have: $$\triangle ABC\cong \triangle FCB.$$

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 $$ADE$$ 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.