如图已知d是 abc中的边bc上的一点∠bad=∠c∠abc的平分线交边ac于e交ad于f那么下列结论中错误的是 A Bac Sketch Coloring Page

如图已知d是 abc中的边bc上的一点∠bad=∠c∠abc的平分线交边ac于e交ad于f那么下列结论中错误的是 A Bac Template

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Take a peek at some of the sketches created by our users, are you a sketchite?

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anonymous

anonymous
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anonymous

anonymous
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anonymous

anonymous
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anonymous

anonymous
sketch #3280 Adventure Time by Alina Danielle

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anonymous

anonymous
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anonymous

anonymous
sketch #1846 zay

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anonymous

anonymous
sketch #5212 Katy Perry by Shahid Khokhar

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anonymous

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sketch #3779 RIP Hugo Chavez Joey Hyuuga

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anonymous

anonymous
sketch #4900 Moonlight by Abdelwahab Haik

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anonymous

anonymous
sketch #5308 Taylor Swift by Gabi Ferrari

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anonymous

anonymous
sketch #3538

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Laura Agapi Asnicar
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anonymous

anonymous
sketch #3208 ELETR by Raimundo Santos

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anonymous

anonymous
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anonymous

anonymous
sketch #5139 Sick Sick Six Urs Stefan

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anonymous

anonymous
sketch #4979 Luffy

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anonymous

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ShopRasta Roots
sketch #2787 Spongebob by ShopRasta Roots

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anonymous

anonymous
sketch #4012 Owl sketch

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anonymous

anonymous
sketch #3036 Wiz Khalifa by Malyi Link

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anonymous

anonymous
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sketch #2479 C 'est qui ? Un rasta qui a manger trop de poulpe farcis. ? Un vieux chanteur de blue grass ? Najdica tomktoo ? A vous de me dire...

Pascal Quéru
sketch #2479 C 'est qui ? Un rasta qui a manger trop de poulpe farcis. ? Un vieux chanteur de blue grass ? Najdica tomktoo ? A vous de me dire...

sketch #4995

anonymous

anonymous
sketch #4995

tomski如图已知d是 abc中的边bc上的一点∠bad=∠c∠abc的平分线交边ac于e交ad于f那么下列结论中错误的是 a coloring

如图已知D是 ABC中的边BC上的一点∠BAD=∠C∠ABC的平分线交边AC于E交AD于F那么下列结论中错误的是 A BAC sketch template

如图已知在 ABC中AD是BC边上的中线E是AD上一点且BE=AC延长BE交AC于点F。求证AF=EF。百度教育 sketch template

如图已知∠BAC=70°D是 ABC的边BC上的一点且∠CAD=∠C∠ADB=80°．求∠B的度数．题目和参考答案——青夏教育精英家教�—— sketch template

如图已知D是 ABC的BC边的中点DE⊥ACDF⊥AB垂足分别为EF且DF=DE则 ABC是等腰三角形吗请说明理由．——青夏 sketch template

题目如图已知 ABC是边长为3的等边三角形点D是边BC上的一点且BD＝1以AD为边作等边 ADE过点E作EF∥BC交AC于点F sketch template

已知D是 ABC的边BC上的一点点B和C到AD的距离相等那么线段AD是 ABC的 A．BC的垂直平分线B．角平分线C．中线D．高线 sketch template

如图8 7已知 ABC中D是BC上一点DE⊥ABDF⊥AC垂足分别为EF如果DE=DF∠BAC=60°AD=20 Cm那么 sketch template

题目如图已知正方形DEFG的顶点DE在 ABC的边BC上顶点GF分别在边ABAC上．如果BC=4 ABC的面积是6那么这个正 sketch template

如图，在 ABC中，点D是BC上一点，E是AD上一点，EB=EC，∠ABE=∠ACE。求证：AD⊥BC sketch template

如图，已知D，E分别是 ABC的边AC，BC上的点，平面α经过D，E两点． 1作直线AB与平面α的交点P； 2求证：D，E，P三点共线 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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

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

题目在 ABC中∠BAC＝90°点D是BC上一点将 ABD沿AD翻折后得到 AED边AE交射线BC于点F．友情提醒翻折前后的 sketch template

3．已知如图7D是 ABC的边AB上一点AB∥FCDF交AC于点EDE＝EF 求证AE＝CE——青夏教育精英家教�—— sketch template

如图DE是 ABC中BC边的两个三等分点F是AC的中点AD与EF交于点O则S OAES ABC= ．题目和参考答案——青夏教育精英 sketch template

百度教育 sketch template

如图D是等边三角形ABC的边AB上一点且ADDB=12现将 ABC折叠使点C与点D重合折痕为EF点EF分别在AC和BC上且 sketch template

如图1，D、E、F分别是 ABC各边上的点，四边形ADEF是平行四边形，有三个选项：�D是AB的中点，�E是BC的中点，�F是AC的中点． sketch template

如图，已知以 ABC的边AB、AC分别�外作等腰Rt ABD与等腰Rt ACE，其中∠BAD=∠CAE=90°，连接BE、CD，BE和CD相交 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

如图，Rt ABC中，∠C＝90°，AC＝BC，点D、E分别是边AB、AC上的点，把 ADE沿DE折叠，点A恰好落在BC上的点F处，若点F为 sketch template

如图在ΔABC中D是BC的中点连接AD并延长到E使DE=AD连接CE试判断AB与EC的位置关系和数量关系并说明理由——青夏教育 sketch template

6．如图已知DE分别是 ABC的ABAC边上一点DE//BC且那么ADAB等于 A．1 B．18 C．19——青夏 sketch template

已知如图在 ABC中D是AB边上的一点且BD=2ADCD=10 Sin∠BCD=则BC边上的高AE 的长为 A．45 B．6 sketch template

已知如图⊿ABC中∠A = 90AB =ACD是BC边上的中点EF分别是ABAC上的点且BE AF求证ED⊥FD sketch template

如图ABC中AB=6AC=8BC=10DE分别是ACAB的中点则以DE为直径的圆与BC的位置关系是 A 相交 B 相切 sketch template

如图等边 ABC中点DE分别是边ABAC的中点CDBE交于点O则∠BOC的度数是120度．题目和参考答案——青夏教育精英家教�—— sketch template

如图已知 ABC中∠C=90°点E在边AB上请过点E作一条直线与 ABC的其他边相交于一点D使得以点AED为顶点的三角形与 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AB＞ACD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已知 ABC的内角∠A=α°分别作内角∠ABC与外角∠ACD的平分线两条平分线交于点A1得∠A1∠A1BC和∠A1CD的平分线交 sketch template

百度教育 sketch template

如图在等边 ABC中D是AC的中点延长BC到点E使CE=CDAB=10cm．1求BE的长2 BDE是什么三角形为什么 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

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