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总结帝笔记—初三寒假班第三讲

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对于该题,知识点为:平行线分线段成比例和
相似。

就记住:“对应”;“交叉可互换”

A选项: \frac{AD}{AN} = \frac{AB}{AM}

B选项: \frac{BD}{AD} = \frac{MN}{AN} = \frac{CE}{AE}

C. 正确

D选项: \frac{DN}{BM} = \frac{AN}{AM} = \frac{NE}{MC}

延伸: (纯自创)

已知: DE \parallel BC

AN = 2NM

\frac{S_{\triangle DEM}}{S_{\triangle ABC}} = ?

Diagram showing triangle ABC with points D on AB and E on AC such that DE is parallel to BC. M is a point on BC. Line segment AM intersects DE at N. Line segments DM and EM are also drawn, forming the quadrilateral ADEM.
Diagram showing triangle ABC with points D on AB and E on AC such that DE is parallel to BC. M is a point on BC. Line segment AM intersects DE at N. Line segments DM and EM are also drawn, forming the quadrilateral ADEM.

看到(面积) (平行) (比例) \Rightarrow 直接锁定: 相似,

(相似比) ^2 = 面积比 \Rightarrow 这就是对知识点的敏感度

干掉它!

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