In the parallelogram $ABCD$ $A$ is the small angle and $BC=2AB$. We draw the height $CE$ to $AB$.We draw a line from $C$ and $E$ to $M$. if $M$ is the midpoint of $AD$ find the measure of $DME$ in terms of $AEM$.
I really cannot do anything special but I have drawn a picture May helps.
Show that each of the two triangles $\triangle CME$ and $\triangle CDM$ is isoceles. This gives you several angle equalities, which (added together correctly) will give $\angle DME$ as a multiple of $\angle AEM$.
A first step that might help is to let $N$ be the midpoint of segment $CE$. Then $MN$ is parallel to both $AE$ and $CD$, and you will find that $\triangle CNM$ and $\triangle ENM$ are two congruent right triangles.