(1)求证: $A F$ 是 $\odot O$ 的切线;
(2)求证: $A B^2-B E^2=B E \cdot E C$;
(3) 如图 2, 若点 $G$ 是 $\triangle A C D$ 的内心, $B C \cdot B E=64$, 求 $B G$ 的长.
【答案】 （1）证明 连接$OA$
\begin{aligned} & \because A C=A B \quad \therefore \widehat{AC}=\widehat{A B} \\ & \therefore O A \perp B C \\ & \because A C=A F \\ & \therefore \angle A C D=\angle C A F+\angle F=2 \angle F \\ & \because A C=A B \\ & \therefore \angle A C B=\angle D \\ & \because C D \| A B \\ & \therefore \angle A B C=\angle B C D \quad \because \angle B=\angle D \\ & \therefore \angle A C B=\angle B C D \\ & \therefore \angle A C D=2 \angle B C D \\ & \therefore \angle B C D=\angle F \\ & \therefore C B \| A F \\ & \therefore O A \perp A F \\ & \because \text { A在圆O上} \\ & \therefore \text { AF是其切线} \\ \end{aligned}

（2）
\begin{aligned} & \angle B A D=\angle B C D=\angle A C B \\ & \angle B=\angle B \\ \therefore & \triangle A B E \backsim \triangle C B A \end{aligned}
\begin{aligned} \therefore \frac{A B}{B C}=\frac{B E}{A B} & \Rightarrow A B^2=B C \cdot B E=B E(B E+C E) \\ & =B E^2+B E \cdot C E \\ \therefore A B^2-B E^2 & =B E \cdot E C \end{aligned}

(3) 由(2）知AB=8

\begin{aligned} & \therefore \angle B A G=\angle B A D+D A G \\ & \angle B G A=\angle G A C+\angle A C B \\ & \because G \text { 为内心 } \\ & \therefore \angle D A G=\angle G A C \\ & \text { 又 } \because \angle B A D+\angle D A G=\angle G A C+\angle A C B \\ & \angle B A D=\angle A C B \\ & \therefore \angle B A G=\angle B G A \\ & \therefore B G=A B=8 \end{aligned}