Let $ABCD$ be a quadrilateral. Prove that if $\overline{AB}$ is congruent to $\overline{CD}$, and $\angle BCD$ is congruent to $\angle DAB$, then $ABCD$ is a parallelogram.
I am feeling stuck because I can't find any avenue to go unless I can somehow prove an angle bisector exists between a diagonal and the pair of congruent angles.
Take $\Delta ABE$ such that $AB=BC$ and let $D\in AE.$
Now, take $\Delta BCD\cong\Delta DEB$ such that $C$ and $E$ are placed in the same half plane respect to $BD$.
Thus, $$AB=BE=CD$$ and $$\measuredangle DAB=\measuredangle BED=\measuredangle BCD,$$ but $ABCD$ is not parallelogram.