$18 \cdot 68$ As shown in the figure, $A B C D$ is a square, and it is also known that $\triangle A B E$ is an equilateral triangle with point $E$ outside the square $A B C D$. The degree measure of $\angle A E D$ is
(A) 10 .
(B) 12.5 .
(C) 15 .
(D) 20 .
(E) 25 .
[Solution] In $\triangle A D E$, since $\angle D A E=90^{\circ}+60^{\circ}=150^{\circ}$, and $D A=$ $A B=A E$, then
$$
\angle A E D=\left(\frac{180-150}{2}\right)^{\circ}=15^{\circ} .
$$

Therefore, the answer is $(C)$.