12．如图，一块三角形铁片$ABC$，已知$AB = 4$，$AC = 4\sqrt{3}$，$\angle BAC = \frac{5\pi}{6}$，现在这块铁片中间发现一个小洞，记为点$D$，$AD = 1$，$\angle BAD = \frac{\pi}{6}$，如果过点$D$作一条直线分别交$AB$，$AC$于点$E$，$F$，并沿直线$EF$裁掉$\triangle AEF$，则剩下的四边形$EFBC$面积的最大值为（ ）
A．$3\sqrt{3}$ B．$2\sqrt{3}$ C．$\sqrt{6}$ D．$\sqrt{3}$
解：以$AC$所在直线为$x$轴，$A$点为坐标原点建立平面直角坐标系
依题可知$B\left( -2\sqrt{3},2 \right)$，$D\left( \frac{1}{2},\frac{\sqrt{3}}{2} \right)$，$AB$所在直线方程为$y = - \frac{\sqrt{3}}{3}x$
设$EF$所在直线方程为$y - \frac{\sqrt{3}}{2} = k\left( x + \frac{1}{2} \right)$，则$0 - \frac{\sqrt{3}}{2} = k\left( x_{F} + \frac{1}{2} \right)$
$\therefore x_{F} = - \frac{k + \sqrt{3}}{2k}$
联立$\begin{cases} y - \frac{\sqrt{3}}{2} = k\left( x + \frac{1}{2} \right) \\ y = - \frac{\sqrt{3}}{3}x \end{cases}$，得$y_{E} = \frac{k + \sqrt{3}}{2(1 + \sqrt{3}k)}$
$\therefore S_{\triangle AEF} = \frac{1}{2} \cdot \left( - \frac{k + \sqrt{3}}{2k} \right) \cdot \frac{k + \sqrt{3}}{2(1 + \sqrt{3}k)} = - \frac{(k + \sqrt{3})^{2}}{8k(1 + \sqrt{3}k)}$
$\therefore (1 + 8\sqrt{3}s)k^{2} + (2\sqrt{3} + 8s)k + 3 = 0$
$\therefore \Delta = (2\sqrt{3} + 8s)^{2} - 4 \cdot (1 + 8\sqrt{3}s) \cdot 3 \geq 0$
$\therefore s \geq \sqrt{3}$
$\therefore S_{四边形EFBC} = S_{\triangle ABC} - s \leq \frac{1}{2} \cdot 4 \cdot 4\sqrt{3} \cdot \frac{1}{2} - \sqrt{3} = 3\sqrt{3}$ 选 A