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

已知在三角形abc中知识要点北京爱智康 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

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

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

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

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

已知在等边三角形ABC中DE分别为ABAC上的点且BD=AEEB与CD相交于点OEF⊥CD于点F．求证OE=2OF．题目和参考 sketch template

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

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

百度教育 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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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

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

26．如图已知 ABC中DE分别是ACAB上的点BD与CE相交于点O给出下列三个条件 �∠EBO=∠DCO �∠BEO=∠CDO sketch template

25．如图在 ABC中 DE分别是ACAB上的点 BD与CE交于点O 给出下列三个条件�∠EBO＝∠DCO�∠BEO＝∠CDO 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

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

如图在三角形纸片ABC中AB=6BC=8AC=4．沿虚线剪下的涂色部分的三角形与 ABC相似的是ABCD 题目和参考答案——青 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

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

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

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

如图线段AB经过圆心O交⊙O于点AC∠A=∠B=30º边BD交⊙O于点D试说明BD是⊙O的切线——青夏教育精英家教�—— 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

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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