In parallelogram ABCD, BAD = 110^. E is a point on AD such that AE = AB. Calculate CED.

In a parallelogram, opposite sides are equal and consecutive angles are supplementary.
Given $\angle BAD = 110^\circ$.
Since E is on AD, $\angle BAE = \angle BAD = 110^\circ$.

We are given that $AE = AB$. Since $AB = 50$ mm, then $AE = 50$ mm.
Because $AE = AB$, $\triangle ABE$ is an isosceles triangle.
The base angles of an isosceles triangle are equal