The problem is: Given parallelogram ABCD with diagonals $\overline{\rm AC}$ and $\overline{\rm BD}$ intersecting at E. If m $\angle\ AEB$ = 60 degrees, m $\angle\ CAB$ = 30 degrees, and AB = 18, find AD.
This problem seems simple, and I presume that you use the properties of special right triangles; however, I can't seem to get the right answer. The answer is 6 $ \sqrt \ 21$. Thank you!
Since $\measuredangle ABE=90^{\circ}$, we obtain $$BE=18\tan30^{\circ}=6\sqrt3,$$ which gives $$BD=12\sqrt3.$$ Thus, $$AD=\sqrt{18^2+\left(12\sqrt3\right)^2}=\sqrt{756}=6\sqrt{21}.$$