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

已知在三角形abc中知识要点北京爱智康 sketch template

有哪些几何题添加的辅助线令人拍案叫绝？ 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

Java编程证明：三角形A2B2C2面积是三角形ABC面积的七分之一 已知一个正三角形abc其内分点a1、b1、c1使得2ac1 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

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

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

百度教育 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

已知在 ABC中∠A＝2∠B 10°∠B＝∠C 20°求三角形的各内角的度数——青夏教育精英家教�—— 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

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

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

百度教育 sketch template

在等边三角形ABC中，点D是BC的中点，点E、F分别是边AB、AC（含线段AB、AC的端点）上的动点，且∠EDF=120°，小明和小慧对这个 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

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

百度教育 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

【初步感知�如图1，已知 ABC为等边三角形，点D为边BC上一动点点D不与点B，点C重合以AD为边�右侧作等边 ADE，连接CE求证：【类比 sketch template

百度教育 sketch template

如图，在 ABC中，AB=10，AC=6，则BC边上的中线AD的取值范围是百度教育 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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

《新手�自学基本功法》 知乎 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

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