$17 \cdot 77$ As shown, in the right-angled $\triangle A B C$ with $A$ as the right-angle vertex, $\angle B=60^{\circ}$, $A D \perp B C, A E$ bisects the interior angle $\angle B A C, A B=a$, then the length of $D E$ is
(A) $(\sqrt{3}-1) a$.
(B) $(\sqrt{3}+1) a$.
(C) $\left(\sqrt{3}-\frac{3}{2}\right) a$.
(D) $\left(\frac{5}{2}-\sqrt{3}\right) a$.
(E) $\left(\sqrt{3}+\frac{1}{2}\right) a$.
[Solution] From the given, we know $\angle C=30^{\circ}$, so from $A B=a$, we get $B C=2 a$, thus $\quad A C=\sqrt{(2 a)^{2}-a^{2}}=\sqrt{3} a$.
Also, since $A D \perp B C$, we know $\triangle A B D$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, then
$$
B D=\frac{1}{2} A B=\frac{a}{2} \text {. }
$$

Furthermore, $A E$ bisects $\angle B A C$, so $\frac{B E}{E C}=\frac{A B}{A C}=\frac{a}{\sqrt{3} a}=\frac{1}{\sqrt{3}}$, thus
$$
B E=2 a \cdot \frac{1}{1+\sqrt{3}}=\frac{2 a}{1+\sqrt{3}}=(\sqrt{3}-1) a .
$$

Therefore, $D E=B E-B D=(\sqrt{3}-1) a-\frac{a}{2}=\left(\sqrt{3}-\frac{3}{2}\right) a$. Hence, the answer is $(C)$.