已知：在三角形abc中，ab=ac角bac=20度，d点在ab上且ad=bc求角acd的度数知乎 Sketch Coloring Page

已知：在三角形abc中，ab=ac角bac=20度，d点在ab上且ad=bc求角acd的度数知乎 Template

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tomski已知：在三角形abc中，ab=ac角bac=20度，d点在ab上且ad=bc求角acd的度数知乎 coloring

已知在三角形abc中知识要点北京爱智康 sketch template

已知：在三角形ABC中，AB=AC角BAC=20度，D点在AB上且AD=BC求角ACD的度数 知乎 sketch template

已知：在三角形ABC中，AB=AC角BAC=20度，D点在AB上且AD=BC求角ACD的度数 知乎 sketch template

【中考提升�初中数学平面几何压轴题6大模型及解题方法 知乎 sketch template

如图，在三角形 ABC 中， D 是 BC 的中点， E 、 F AC 的三等分点。已知三角形 的面积是 108 平方厘米。求三角 sketch template

三角形中线的定理？ 知乎 sketch template

如图，在三角形ABC中，D、E、F分别是BC、CA、AB上的点，AD、BE、CF相交于一点O，这样三角形ABC被分成6个三角形，其中4个三角形 sketch template

已知：在三角形ABC中，AB=AC角BAC=20度，D点在AB上且AD=BC求角ACD的度数 知乎 sketch template

如图等腰三角形ABC中AB=AC顶角∠A=20°在边AB上取一点D使AD=BC求∠BDC的度数 题目和参考答案——青夏教育精英家教�—— sketch template

题目1操作如图在已知内角度数的三个三角形中请用直尺从某一顶点画一条线段把原三角形分割成两个等腰三角形并在图中标注相应的角的 sketch template

三角形“四心”有关性质及证明（内心篇） 知乎 sketch template

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如图已知在 ABC中AD是BC边上的中线E是AD上一点且BE=AC延长BE交AC于点F。求证AF=EF。百度教育 sketch template

已知三角形ABC中AB≠AC求证∠B≠∠C要用反证法证明此�题第一步应假设——青夏教育精英家教�—— sketch template

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在三角形ABC中点O是AC边上一个动点过点O作MN∥BC设MN交∠BCA的平分线于点E交 ABC的外角∠ACD平分线于点E．1求证 sketch template

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如图三角形ABC中AB=3AC=6∠BAC=60°D为BC中点E为中线AD的中点．1试用�量AB和AC表示AD2求中线AD sketch template

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如图在三角形ABC中BC=8厘米AD=6厘米EF分别为AB和AC的中点．那么三角形EBF的面积是66平方厘米．——青夏教育精英家教�—— sketch template

19．已知如图7等腰三角形ABC中AB=AC=4若以AB为直径的⊙O与BC相交于点DDE∥ABDE与AC相交于点E则DE sketch template

记 ABC的内角ABC的对边分别为abc已知 =ac点D在边AC 上BDsin∠ABC=asinC1证明BD = B sketch template

已知 ABC是等边三角形以BC为直径的半圆O与边AB相交于点DDE⊥AC垂足为点E．1判断DE与⊙O的位置关系并证明你的结论2 sketch template

百度教育 sketch template

如图在不等边 ABC中AB＞ACAC≠BC过AC上一点D作一条直线使截得的三角形与原三角形相似这样的直线可作 A2条 B3条 sketch template

如图在不等边 ABC中AB＞ACAC≠BC过AC上一点D作一条直线使截得的三角形与原三角形相似这样的直线可作 A2条 B3条 sketch template

如图在三角形ABC中AB＞ACDE分别是ABAC上的点 ADE沿线段DE翻折使点A落在边BC上记为A′．若四边形ADA′E是 sketch template

在三角形ABC中∠B＝60°BA＝24cmBC＝16cm现有动点P从点A出�沿射线AB�点B方�运动动点Q从点C出�沿射线CB sketch template

如图已知在 ABC中AB=ACD为BC边的中点过点D作DE⊥ABDF⊥AC垂足分别为EF．1求证 BED≌ CFD2 sketch template

如图在 ABC中已知点DE分别在ABAC上BE与CD相交于点O依据下列各个选项中所列举的条件不能说明AB=AC的是 A．BE sketch template

如图 ABC中AB=AC．∠A=36°AB的中垂线DE交AC于D交AB于E下述结论AD=BD=BC3 BDC的�长等于AB sketch template

百度教育 sketch template

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