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，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

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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

【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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在三角形ABC中角ABC的对边分别为abc已知sinAsinB SinBsinC Co ZOL问答 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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百度教育 sketch template

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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

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已知如图在 ABC中AB=ACD是AB上一点过D作DE⊥BC于E并与CA的延长线相交于F求证AD=AF——青夏教育精英家教�—— 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

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

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

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 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

在Rt ABC中∠ACB=90°∠BAC=30°BC=6．I如图�将线段CA绕点C顺时针旋转30°所得到与AB交于点M则CM的 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

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