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49 lines
1.7 KiB
49 lines
1.7 KiB
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### 一、第一问
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$\because$ 圆中同弧所对的圆周角相等
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$\therefore$ $\angle AED=\angle ABD$ ①
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连接$AD$, $\because \angle ADB$是直径$AB$所对的圆周角，$\therefore \angle ADB=90^{\circ}$
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又 $\because D$是$BC$中点，所以$\triangle ADB \cong \triangle ADC$
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$\therefore \angle ABD = \angle ACB$ ②
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联立 ① ②, $\therefore \angle AED=\angle C$
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### 二、第二问
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$\because \angle AED=55^{\circ}$
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$\therefore \angle ABC=\angle C=55^{\circ}$
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$\therefore \angle BAC=180^{\circ}-55^{\circ}-55^{\circ}=70^{\circ}$
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根据 <font color='blue' size=4><b>圆的内接四边形对角互补</b></font>,$\therefore \angle BDF=110^{\circ}$
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### 三、第三问
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看到线段乘积，考虑找到相似三角形
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$$\large \frac{EG}{?}=\frac{?}{ED}$$
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考虑$\triangle EAG \sim \triangle EAD$
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如何证明呢？
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有一个公共角$\angle AED$
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还需要再找一个角：
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因为$E$是 $\overset{{\frown}}{AB}$ 的中点，
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$\therefore \angle ADE=\angle BAE$
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$\therefore \triangle EAG \sim \triangle EAD$
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$$\large \frac{EG}{AE}=\frac{AE}{ED}$$
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$\Rightarrow$ $EG*ED=AE^2$
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$AE$长度如何求解呢？
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$\because \angle AFD+\angle ABD=180^{\circ}$
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$\because \angle AFD+\angle CFD=180^{\circ}$
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$\therefore \angle ABD=\angle CFD$
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因为有第一问的结论，所以$\triangle DCF$是等腰三角形，$CD=DF=BD=4$
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$\because cos\angle ABD=\frac{2}{3}$
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$\large \therefore \frac{BD}{AB}=\frac{2}{3}$
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$\therefore AB=6$
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$\because \triangle ABE$是等腰直角三角形
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$\therefore AE=3\sqrt{2}$
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$\therefore EG*ED=(3\sqrt{2})^2=18$ |