Pls help cphill

Since ∠ACB=90∘, we have ∠ACE=30∘. Also, since △DEF is equilateral, ∠DEF=60∘. Then ∠AED=60∘−30∘=30∘. Furthermore, △AED is isosceles since AD=AE, so ∠ADE=30∘. Since the sum of the angles in a triangle is 180∘, we have $$\begin{align*} \angle AED + \angle ADE + \angle EAD &= 180^\circ \ 30^\circ + 30^\circ + \angle EAD &= 180^\circ \ \angle EAD &= 120^\circ. \end{align*}$$ By the Law of Sines in △ADE, we have $$\begin{align*} \frac{AD}{\sin \angle EAD} &= \frac{AE}{\sin \angle ADE} \ \frac{AE}{5} &= \frac{AE}{\sin 30^\circ} \ 5 &= \sin 30^\circ \ AE &= \frac{5}{\sin 30^\circ} \ AE &= \boxed{\frac{10}{\sqrt{3}}}. \end{align*}$$

So the answer is AE = 10/sqrt(3).