Find point $E$ on $CD$ of parallelogram $ABCD$ such that $\angle AEB = \angle BEC$

Question

Find point $E$ on $CD$ of parallelogram $ABCD$ such that $\angle AEB = \angle BEC$

Shape is supposed to look something like this.

Answer 1

Since $AB || CD$ we have $<BEC =<EBA$ (wherever $E$ is on $CD$).

If we also have this common angle equal to $<AEB$ then $\Delta AEB$ is isosceles, with $AE=AB$.

therefore: Draw the circle centered at $A$ with radius $AB$ and intersect it with the line $CD$.