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，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

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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

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，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

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三角形“四心”有关性质及证明（内心篇） 知乎 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

三角形中线的定理？ 知乎 sketch template

百度教育 sketch template

如图已知在等腰 ABC中AB=AC顶角∠A=20°在AB边上取D点使得AD=BC则∠BDC的度数等于题目和参考答案——青夏教育 sketch template

百度教育 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

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

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

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如图在 ABC中∠BAC=90°AB=ACBE平分∠ABCCE⊥BE求证CE=12BD．题目和参考答案——青夏教育精英家教�—— sketch template

百度教育 sketch template

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

在三角形ABC中如果A点的位置用数对表示为45B点的位置用数对表示为42C点的位置用数对表示为32那么三角形ABC sketch template

百度教育 sketch template

百度教育 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

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

在等腰三角形ABC中AB=ACBECD分别是底角的平分线DE∥BC图中等腰三角形的个数有 A3个 B4个 C5个 D6 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在三角形ABC中AB=AC点D在BC延长线上点E在AC上连接AD连接BE并延长交AD于F连接FC已知∠EBC=∠D．1 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

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 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

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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

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