(本小题 8.0 分)

如图, 四边形 $A B C D$ 内接于 $\odot O, A B=A C, B D \perp A C$, 垂足为点 $E$, 点 $F$ 在 $B D$ 的延长线上,且 $D F=D C$, 连接 $A F, C F$, 若 $A F=10, B C=4 \sqrt{5}$, 求 $\tan \angle B A D$ 的值.

<ImageHere>
解:

$\because A B=A C$,

$\therefore \widehat{A B}=\overparen{A C}, \angle A B C=\angle A C B$,

$\therefore \angle A B C=\angle A D B, \angle A B C=\frac{1}{2}\left(180^{\circ}-\angle B A C\right)=90^{\circ}-\frac{1}{2} \angle B A C$,

$\because B D \perp A C$,

$\therefore \angle A D B=90^{\circ}-\angle C A D$,

$\therefore \frac{1}{2} \angle B A C=\angle C A D$,

$\therefore \angle C A D=\angle F B C=\frac{1}{2} \angle B A C$,
$\because D F=D C$,

$\therefore \angle B F C=\frac{1}{2} \angle B D C=\frac{1}{2} \angle B A C=\angle F B C$,

$\therefore C B=C F$,

又 $B D \perp A C$,

$\therefore A C$ 是线段 $B F$ 的中垂线,

$\therefore A B=A F=10, A C=10$.

又 $B C=4 \sqrt{5}$, 设 $A E=x, C E=10-x$,

则 $A B^{2}-A E^{2}=B C^{2}-C E^{2}$, 即 $100-x^{2}=80-(10-x)^{2}$, 解得 $x=6$,

$\therefore A E=6, B E=8, C E=4$,

$\because \triangle D E C \sim \triangle A E B$,

$\therefore D E=\frac{A E \cdot C E}{B E}=\frac{6 \times 4}{8}=3$,

作 $D H \perp A B$, 垂足为点 $H$,

<ImageHere>

$\because \angle D B H=\angle A B E, \angle B H D=\angle A E B=90^{\circ}$,

$\therefore \triangle D B H \sim \triangle A B E$,

$\because \frac{D B}{A B}=\frac{B H}{B E}=\frac{D H}{A E}$,

$\therefore \frac{11}{10}=\frac{B H}{8}=\frac{D H}{6}$,

$\therefore B H=\frac{44}{5}, D H=\frac{33}{5}$,

$\therefore A H=A B-B H=10-\frac{44}{5}=\frac{6}{5}$,
$\therefore \tan \angle B A D=\frac{D H}{A H}=\frac{\frac{33}{5}}{6}=\frac{11}{2}$,

【解析】见答案