8．如图所示o是等边三角形abc的中心d是三角形中的任意一点如果作矢量dadbdc分别表示三个力三个力的方�如图中箭头所示则这三 Sketch Coloring Page

8．如图所示o是等边三角形abc的中心d是三角形中的任意一点如果作矢量dadbdc分别表示三个力三个力的方�如图中箭头所示则这三 Template

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sketch #1846 zay

anonymous

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anonymous

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BaGaz Anggara
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sketch #3707 The Thinker

anonymous

anonymous
sketch #3707 The Thinker

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anonymous

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anonymous

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sketch #2715 Angelina

Na Ruro
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anonymous

anonymous
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sketch #4503 Naruto Biju Mode by Jessie Lightwood

anonymous

anonymous
sketch #4503 Naruto Biju Mode by Jessie Lightwood

sketch #2476 Asuna from SAO by sketchmaster

sketchmaster
sketch #2476 Asuna from SAO by sketchmaster

sketch #5178 Mewtwo and Transformation by Riley Hart

anonymous

anonymous
sketch #5178 Mewtwo and Transformation by Riley Hart

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Chrystal Smith
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Laura Agapi Asnicar
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anonymous

anonymous
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sketch #4349 Cícero Николай Герасимов

anonymous

anonymous
sketch #4349 Cícero Николай Герасимов

sketch #2723 by Sam Torres

Sam Torres
sketch #2723 by Sam Torres

sketch #5271

anonymous

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sketch #5268 Harley Quinn by 조바다

anonymous

anonymous
sketch #5268 Harley Quinn by 조바다

sketch #1079

anonymous

anonymous
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tomski8．如图所示o是等边三角形abc的中心d是三角形中的任意一点如果作矢量dadbdc分别表示三个力三个力的方�如图中箭头所示则这三 coloring

等边三角形八等分图片千图� sketch template

如图所示在等边 ABC中点DE分别在边BCAB上且BD=AEAD与CE交于点F则∠DFC的度数为 A．60°B．45°C sketch template

如图， ABC是等边三角形，D是BC边上一点，且满足∠ADE=60∘，DE交等边三角形ABC的外角平分线CE所在直线于点E，试探究AD与DE的 sketch template

题目如图所示边长为2的等边三角形ABC中D点在边BC上运动不与BC重合点E在边AB的延长线上点F在边AC的延长线上AD sketch template

8．如图所示O是等边三角形ABC的中心D是三角形中的任意一点如果作矢量DADBDC分别表示三个力三个力的方�如图中箭头所示则这三 sketch template

如图 ABC为等边三角形以边BC为直径的半圆与边ABAC分别交于DF两点过点D作DE⊥AC垂足为点E．1判断DE与⊙O的位置 sketch template

如图，已知 ABC是等边三角形，延长BA至E，延长BC至D，使AE=BD，求证：EC=ED．百度教育 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1等边三角形ABC的边长为8点P由点B开始沿BC以每秒1个单位长的速度作匀速运动到点C后停止运动点Q由点C开始沿C A B以每 sketch template

如图点O是等边 ABC内一点D是 ABC外的一点∠AOB=110°∠BOC=α BOC≌ ADC∠OCD=60°连接OD．1 sketch template

在等边三角形ABC中，点D、E分别在AB、BC边上，连结CD，AE交于点P，且CD=AE，∠ BCD=20°，则∠ APD的度数为百度教育 sketch template

如图3 ABC是等边三角形在ACBC边上各取一点PQ使AP=CQ AQ BP交于点O则∠QOB= ．——青夏教育精英家教�—— sketch template

已知 ABC为等边三角形，边长为8，点D，E分别是边AB，BC上的动点，以DE为边作等边三角形DEF1如图1，若点F落在边AC上�求证 sketch template

10．如图所示已知 ABC和 DCE均是等边三角形点BCE在同一条直线上AE与BD交于点OAE与CD交于点GAC与BD交于点F sketch template

如图所示，在 ABC中，∠B=60°，AB=AC，点D、E分别在BC、AB上，且BD=AE，AD与CE交于点F．（1）求证： ABC是等边三角 sketch template

如图所示 ABC是等边三角形D点是AC的中点延长BC到E使CE=CD 1用尺规作图的方法过D点作DM⊥BE 垂足是M不 sketch template

如图所示，边长为2的等边三角形中，D点在边上运动（不与B、C重合），点E在边的延长线上，点F在边的延长线上，．点D在边上从B至C的运动过程 sketch template

1如图1所示若P为等边 ABC内一点∠BPC=150° BPP′是等边三角形求证PC2 PB2=PA22如图2所示点P为 sketch template

如图， ABC和 DEF是两个全等的等腰直角三角形，∠ BAC=∠ EDF=90°， DEF的顶点E与 ABC的斜边BC的中点重合，将 DEF sketch template

10．如图所示已知 ABC和 DCE均是等边三角形点BCE在同一条直线上AE与BD交于点OAE与CD交于点GAC与BD交于点F sketch template

如图所示已知等边三角形ABC的边长为1按图中所示的规律用2011个这样的等边三角形镶嵌而成的四边形的�长是 ．题目和参考答案——青夏 sketch template

百度教育 sketch template

1如图1所示在等边 ABC中点D是AB边上的动点以CD为一边�上作等边 EDC连接AE求证AE∥BC中等边 ABC的形状改 sketch template

1如图1点O是线段AD的中点分别以AO和DO为边在线段AD的同侧作等边三角形OAB和等边三角形OCD连接AC和BD相交于点E连接 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等边三角形ABC的边长为4P是三角形内角任意一点过点P作三边的垂线PDPEPF垂足分别为DEF．则PD PE PF sketch template

如图⊿ABC是等边三角形点DE分别在BCAC上且BD=CEAD与BE相交于点F 1试说明⊿ABD≌⊿BCE 2⊿AEF sketch template

如图O是等边 ABC内一点OA=3OB=4OC=5将线段BO绕点B逆时针旋转60°得到线段BO′．1求点O与O′的距离2 sketch template

如图所示已知 ABC和 DCE都是等边三角形图中的三角形可以通过旋转相互得到的是A ACE和 BCDB ABF和 CFDC ABC sketch template

如图1⊙O是等边三角形ABC的外接圆P是⊙O上的一个点．1则∠APC=60°2试证明PA PB=PC3如图2过点A作⊙ sketch template

如图所示已知 ABC和 DCE都是等边三角形图中的三角形可以通过旋转相互得到的是A ACE和 BCDB ABF和 CFDC ABC sketch template

如图所示P是等边 ABC内部一点∠APB∠BPC∠CPA的大小之比是5∶6∶7求以PAPBPC的长为边的三角形的三个角的大小之比 sketch template

12．如图所示 ABC中AB=AC∠BAC=4∠BABAC的垂直平分线分别交BC于点DE．认真分析下列说法 � BAE和 CAD sketch template

如图所示直线AD和BC相交于OAB∥CD∠AOC=95°∠B=50° ∠A=A50B55C40D45——青夏教育精英 sketch template

如图所示已知 ABC和 BDE都是等边三角形且ABD三点共线．下列结论�AE=CD�BF=BG�HB平分∠AHD�∠AHC sketch template

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