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

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、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

百度教育 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

已知在三角形abc中知识要点北京爱智康 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

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

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

$�们定义等腰三角形中底边与腰的比叫做底角的邻对在 ABC中AB=AC底角B的邻对记作canB这时canB=$frac{底边}{腰 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

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

百度教育 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

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

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

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

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

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

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

已知如图在 ABC中AB=AC在AC上取点E在AB的延长线上取点D使BD=EC连接DE交BC于点F．求证DF=EF．题目和 sketch template

已知如图在 ABC中AB=AC点DEF分别在边BCBACA上且BE=CDBD=CF．求证点D在线段EF的垂直平分线上 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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

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

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 ABC中AB＞AC＞BCP为三角形内任意一点连接AP并延长交BC于D．求证1AB AC＞AD BC2AB AC sketch template

百度教育 sketch template

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