![](http://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/2023/04/d69383cd7619884fef6b4b8252f45292.png)

### 一、第一问
$\because$ 圆中同弧所对的圆周角相等
$\therefore$ $\angle AED=\angle ABD$ ①

连接$AD$, $\because \angle ADB$是直径$AB$所对的圆周角，$\therefore \angle ADB=90^{\circ}$
又 $\because D$是$BC$中点，所以$\triangle ADB \cong \triangle ADC$
$\therefore \angle ABD = \angle ACB$ ②

联立 ① ②, $\therefore \angle AED=\angle C$

### 二、第二问
$\because \angle AED=55^{\circ}$
$\therefore \angle ABC=\angle C=55^{\circ}$
$\therefore \angle BAC=180^{\circ}-55^{\circ}-55^{\circ}=70^{\circ}$
根据 <font color='blue' size=4><b>圆的内接四边形对角互补</b></font>,$\therefore \angle BDF=110^{\circ}$

### 三、第三问
看到线段乘积，考虑找到相似三角形
$$\large \frac{EG}{?}=\frac{?}{ED}$$
考虑$\triangle EAG  \sim \triangle EAD$

如何证明呢？
有一个公共角$\angle AED$
还需要再找一个角：
因为$E$是 $\overset{{\frown}}{AB}$ 的中点，
$\therefore \angle ADE=\angle BAE$

$\therefore \triangle EAG  \sim \triangle EAD$


$$\large \frac{EG}{AE}=\frac{AE}{ED}$$
$\Rightarrow$ $EG*ED=AE^2$

$AE$长度如何求解呢？
$\because \angle AFD+\angle ABD=180^{\circ}$
$\because \angle AFD+\angle CFD=180^{\circ}$
$\therefore \angle ABD=\angle CFD$
因为有第一问的结论，所以$\triangle DCF$是等腰三角形，$CD=DF=BD=4$

$\because cos\angle ABD=\frac{2}{3}$
$\large \therefore \frac{BD}{AB}=\frac{2}{3}$
$\therefore AB=6$

$\because \triangle ABE$是等腰直角三角形
$\therefore AE=3\sqrt{2}$

$\therefore EG*ED=(3\sqrt{2})^2=18$