2014 AMC 10A Problem 22

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.