Answer: E

Small Hint:

Construct a helpful point making a $30$-$60$-$90$ triangle

Big Hint:

Show the constructed point is the same as $E$

Solution:

Let $E'$ be the point on $\overline{CD}$ such that $AE'=AB=20$.

Since $AD=10$, triangle $ADE'$ is a $30$-$60$-$90$ triangle, so $\angle DAE'=60^\circ$ and $\angle BAE'=30^\circ$.

Also $AE'=AB$, so triangle $ABE'$ is isosceles. Its vertex angle at $A$ is $30^\circ$, so each base angle is $75^\circ$.

Therefore $\angle CBE'=90^\circ-75^\circ=15^\circ$, so $E'=E$, and $AE=20$.

Thus, E is the correct answer.