27. As shown in Figure $10, B C$ is the diameter of the semicircle $\odot O$, $E F \perp$ $B C$ at point $F, \frac{B F}{F C}=5$. Given $A B=8, A E=2$. Then the length of $A D$ is $(\quad$.
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{1+\sqrt{3}}{2}$
(C) $1+\sqrt{3}$
(D) $1+\sqrt{2}$
27. B.

Connect $B E$. Since $B C$ is the diameter, we know $\angle B E C=90^{\circ}$. Therefore,
$$
B E=\sqrt{A B^{2}-A E^{2}}=\sqrt{8^{2}-2^{2}}=2 \sqrt{15} \text {. }
$$

Also, from Rt $\triangle B F E \backsim \mathrm{Rt} \triangle E F C$, we have
$$
\frac{B E}{E C}=\frac{B F}{E F}=\frac{E F}{F C} \Rightarrow \frac{B E^{2}}{E C^{2}}=\frac{B F}{F C}=5 \Rightarrow E C=2 \sqrt{3} \text {. }
$$

By the secant theorem, we get
$$
A D=\frac{A E(A E+E C)}{A B}=\frac{2(2+2 \sqrt{3})}{8}=\frac{1+\sqrt{3}}{2} .
$$