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

【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、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

三角形中线的定理？ 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

如图在三角形ABC中AB=ACD是BC上一点∠BAD=40°E是AC上一点AD=AE求∠EDC的度数．题目和参考答案——青夏 sketch template

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

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

如图已知 ABC中AB=AC∠BAC=90°点D为BC的中点点EF分别在直线ABAC上运动且始终保持AE=CF．1如图� 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

百度教育 sketch template

百度教育 sketch template

已知 ABC中AB=AC∠BAC=90° ABD为等边三角形E是BD中点AECD相交于O点．1求∠DCB的度数2求证 sketch template

已知如图在等边三角形ABC中MN分别是ABAC的中点D是MN上任意一点CDBD的延长线分别与ABAC交于FE若1CE 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

已知如图在 ABC中AB=AC延长BC到D使BD=2BC连接AD过C作CE⊥BD交AD于点E连接BE交AC于点O．1求证 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

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

如图已知 ABC中AB=ACM是BC的中点DE分别是ABAC边上的点且BD=CE1求证MD=ME．2若D为AB的 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

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

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

定理等腰三角形的两个底角相等．请写已知求证并证明．已知 ABC中AB=AC求证∠B=∠C．证明题目和参考答案——青夏教育精英 sketch template

如图在 ABC中BDCE分别为三角形的两条高交于点O �问图11中有个三角形与相似它们是 �连结DE ADE与 ABC是否相似 sketch template

如图所示，已知边长为4的正三角形ABC中，E，F分别为BC和AC的中点，PA=2，且PA⊥平面ABC，设Q是CE的中点（1）求证：AE∥平面 sketch template

15．如图所示已知等腰直角三角形ABC中∠ACB=90°直线L经过点C AD ⊥LBE⊥L垂足分别为DE． 1证明 ACD sketch template

已知等边三角形ABC中延长AC到F使得ED=DF求证BE=CF——青夏教育精英家教�—— sketch template

如图在Rt ABC中∠C=90°D为AC上一点且DA=DB=5又 DAB的面积为10那么DC的长是 A4 B3 C5 sketch template

已知如图 ABC中∠BAC=90ºAB=ACD为AC的中点AE⊥BD交BC于E试说明∠ADB=∠CDE提示做CG⊥AC sketch template

三角形ABC中，D、E分别是AB边AC边上的点，且AD=AE，BE=CD，求证AB=AC 知乎 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

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

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