20.50 Line segment $AB$ is both the diameter of a circle with radius 1 and a side of equilateral $\triangle ABC$. This circle intersects $AC$ and $BC$ at points $D$ and $E$. The length of $AE$ is
(A) $\frac{3}{2}$.
(B) $\frac{5}{3}$.
(C) $\frac{\sqrt{3}}{2}$.
(D) $\sqrt{3}$.
(E) $\frac{2+\sqrt{3}}{2}$.
(34th American High School Mathematics Examination, 1983)
[Solution] As shown in the figure, since $AB$ is the diameter, $\angle AEB = 90^{\circ}$, so $AE$ is the altitude from $A$ to side $BC$ of the equilateral $\triangle ABC$.

Also, $\angle EAB = \frac{1}{2} \angle CAB = 30^{\circ}$, which means $\triangle ABE$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ right triangle.
Therefore, $AE = AB \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$.
Hence, the answer is $(D)$.