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.