答案：A
連結 $BC$。考慮 $\Delta ABD$ 及 $\Delta ACB$，
$\begin{array}{rcll}
\angle BAD & = & \angle CAB & \text{(公共角)} \\
\angle ACB & = & 90^\circ & \text{(半圓上的圓周角)} \\
\angle ABD & = & 90^\circ & \text{(半徑 $\perp$ 切線)} \\
\therefore \angle ABD & = & \angle ACB \\
\end{array}$
$\begin{array}{rcll}
\angle ADB & = & 180^\circ – \angle BAD – \angle ABD & \text{($\Delta$ 的內角和)} \\
& = & 180^\circ – \angle CAB – \angle ACB & \text{(已證)} \\
& = & \angle ABC & \text{($\Delta$ 的內角和)}
\end{array}$
$\therefore \Delta ABD \sim \Delta ACB$。(A.A.A.)
所以，可得
$\begin{array}{rcl}
\dfrac{AD}{AB} & = & \dfrac{AB}{AC} \\
\dfrac{6}{AB} & = & \dfrac{AB}{4} \\
AB^2 & = & 24 \\
AB & = & \sqrt{24} \\
& = & 2\sqrt{6}
\end{array}$
