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，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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

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已知在 ABC中，AB=AC，点D是边AB上一点，∠BCD=∠A．（1）如图1，试说明CD=CB的理由；（2）如图2，过点B作BE⊥AC，垂足 sketch template

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已知Rt三角形ABC中，AC＝BC，角C＝90度，D为AB边的中点，角EDF＝90度，角EDF绕D点旋转，它的两边分别百度知道 sketch template

连结三角形两边中点的线段叫做三角形的中位线如图�，在 ABC 中，D，E分别是AB，AC的中点，DE 就是 的一条中位线 �们可 sketch template

已知 ABC中AB=AC∠BAC=α DBC为等边三角形．1如图1∠ABD= 2如图2若∠BCE=150°∠ABE=60 sketch template

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

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

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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

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

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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

已知如图在等边三角形ABC中DE分别为BCAC上的点且AE＝CD连结ADBE交于点P作BQ⊥AD垂足为Q．求证BP sketch template

22．已知如图在 ABC中AB=ACAE是角平分线BM平分∠ABC交AE于点M经过BM两点的⊙O交BC于点G交AB于点FFB 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

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

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

如图所示已知三角形ABC中∠BAC为钝角． 1画出点C到AB的垂线段 2过A点画BC的垂线 3点B到AC的距离是多� sketch template

20．已知三角形ABC A C 现将∆ABC平移使点A到位置A’处则点BC对应的坐标分别为 B’C’ ——青夏教育精英家教�—— 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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

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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 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

百度教育 sketch template

百度教育 sketch template

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