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45 lines
1.1 KiB
45 lines
1.1 KiB
2 years ago
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**模型背景：四条线段两两相交**
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**方法：过封闭三角形的顶点作其它线的平行线，构造$A$字和$8$字型相似三角形，用比例解题**
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#### 图($1$)
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先找出相似三角形，再通过相似三角形找出比例关系：
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$$
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\left\{\begin{matrix}
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\triangle AEM \sim \triangle ANC & ①
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\\
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\triangle BNC \sim \triangle BED & ②
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\end{matrix}\right.
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$$
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$\because$ $M$是$AC$中点
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$\therefore \frac{AM}{MC}=\frac{AE}{EN}=1:1$
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又 $\because AE=\frac{1}{4}AB$
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$\therefore AN=BN=2$
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又 $\because S_{\triangle BNC} \sim S_{\triangle BED}$
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$\therefore \frac{BC}{CD}=2$
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#### 图($2$)
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先找出相似三角形，再通过相似三角形找出比例关系：
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$$
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\left\{\begin{matrix}
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\triangle CFM \cong \triangle AEM & ①
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\\
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\triangle CFD \sim \triangle BED & ②
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\end{matrix}\right.
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$$
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$\therefore AE=CF=1,BE=3$
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$\therefore \frac{BC}{CD}=2$
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#### 图(3)
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请同学自己探索
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#### 图(4)
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请同学自己探索
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