如图等边 abc中点d在ab上点e在bc上ad=be．ae和cd相交于点p求证∠cpe=60°．题目和参考答案——青夏教育精英家教�—— Sketch Coloring Page

如图等边 abc中点d在ab上点e在bc上ad=be．ae和cd相交于点p求证∠cpe=60°．题目和参考答案——青夏教育精英家教�—— Template

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sketch #2446 Joker by sketchmaster

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sketch #3572 Hate mondays by Malyi Link

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sketch #3572 Hate mondays by Malyi Link

sketch #2989 Little Chibi by Loai Issa

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sketch #5436

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sketch #5139 Sick Sick Six Urs Stefan

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sketch #5139 Sick Sick Six Urs Stefan

sketch #3244 Winnie the Pooh by Ashok Kumar Chaturvedi

anonymous

anonymous
sketch #3244 Winnie the Pooh by Ashok Kumar Chaturvedi

sketch #4975 Orange by Marcos Roberto Molina

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sketch #4975 Orange by Marcos Roberto Molina

sketch #4983

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sketch #4983

sketch #4710

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sketch #1216

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sketch #4384

Laura Agapi Asnicar
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sketch #5368

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sketch #3367 Ichigo by Szymon Broszczak

anonymous

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sketch #3367 Ichigo by Szymon Broszczak

sketch #5052 Gorillaz by Rafał Dawid

anonymous

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sketch #5052 Gorillaz by Rafał Dawid

sketch #3133 Skrillex by Asfa Sajid

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sketch #3133 Skrillex by Asfa Sajid

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sketch #5199 Catface by Noeli Racioppo

anonymous

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sketch #2899 Snoopy by Rhemzkian Esturco

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anonymous
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tomski如图等边 abc中点d在ab上点e在bc上ad=be．ae和cd相交于点p求证∠cpe=60°．题目和参考答案——青夏教育精英家教�—— coloring

如图所示在等边 ABC中点DE分别在边BCAB上且BD=AEAD与CE交于点F则∠DFC的度数为 A．60°B．45°C sketch template

题目如图等边 ABC的边长为2点D是射线BC上的一个动点以AD为边�右作等边 ADE连结CE1求证 ABD≌ ACE2 sketch template

如图等边三角形ABC的边长为6A、B、A1三点在一条直线上且 ABC≌ A1BC1若D为线段BC1上一动点则AD CD的最小值是 sketch template

题目已知如图 ABC和 CDE均为等边三角形BCD三点在同一条直线上连接线段BEAD交于点F连接CF1求证∠FBC 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

题目如图 ABC是等边三角形AB=点D是边BC上一点点H是线段AD上一点连接BHCH．当∠BHD=60°∠AHC=90°时 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

23． 如图 ABC中点DE分别在边ABAC上连接DE并延长交BC的延长线于点F连接DCBE．若∠BDE ∠BCE=180 sketch template

题目：如图，在三角形ABC中，点D是AB的中点，∠DAC=2∠DCA，∠DCB=30度，求∠B的度数知乎 sketch template

如图等边 ABC中点D在AB上点E在BC上AD=BE．AE和CD相交于点P求证∠CPE=60°． 题目和参考答案——青夏教育精英家教�—— sketch template

题目已知 ABC中∠ACB＝90°AC＝BC．1如图1点D在BC的延长线上连AD过B作BE⊥AD于E交AC于点F．求证 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在 ABC中分别以ABAC为边�外作等腰直角三角形ABE等腰直角三角形ACD其中∠BAE=∠CAD=90°BD与CE相交于点O sketch template

如图等边 ABC中DE分别在ABBC边上且AD=2BE=4连接DE并将线段DE绕点E顺时针旋转60°得到线段EF连接CF取 sketch template

如图在 ABC中内接矩形DEFG其中点DE分别在边ABAC上点GF在边BC上如果EFDE=23BC=8高AH交DE于 sketch template

题目从边长为a的正方形中剪掉一个边长为b的正方形然后将剩余部分拼成一个长方形．1上述操作能验证的等式是 A．a2 2ab B2 sketch template

百度教育 sketch template

百度教育 sketch template

百度教育 sketch template

如图在三角形ABC中点DF在边BC上点E在边AB上点G在边AC上∠CDG=∠B∠1 ∠FEA=180°．证明AD∥EF． 题目 sketch template

如图在平面直角坐标系中把 ABC沿y轴对折后得到 A1B1C1再将 A1B1C1�下平移4个单位长度得到 A2B2C2则 AB1C2的 sketch template

如图在每个小正方形的边长均为1的方格纸中有一个 ABC ABC的三个顶点均与小正方形的顶点重合．1在图中画线段CD使CD∥AB点D sketch template

如图在等边 ABC中D为AC的中点E是AB上一点且则图中和 ADE相似的三角形有A．1个 B．2个 C．3个D．4个——青夏教育精英 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

如图等边 ABC中AB=5点DE分别在BCAC边上将 EDC沿直线DE翻折后点C落在点C′处且点C′在 ABC的外部则图中 sketch template

2012•白下区一模如图直线l经过等边三角形ABC的顶点B在l上取点DE使∠ADB=∠CEB=120°．若AD=2cmCE sketch template

如图在等边 ABC中P是BC上一点D是AC上的一点∠APD=60°BP=1CD=则 ABC的边长是A 3 B 4 C 5 sketch template

百度教育 sketch template

百度教育 sketch template

21．如图在平面直角坐标系中直线AB与Y轴和X轴分别交于点A点8与反比例函数y一罟在第一象限的图象交于点c．过点C作CE上y轴于E过 sketch template

如图，在四边形ABCD中，对角线AC与BD交于点O，已知OA =OC，OB=OD，过点O作EF⊥ BD，分别交AB、DC于点E，F，连接DE sketch template

百度教育 sketch template

如图在 ABC中点D在AB上点E在AC上AD=AE∠ABE=∠ACDBE与CD相交于点F试判断 BFC的形状并说明理由． 题目 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所示在等腰Rt ABC中AC=BC以斜边AB为一边作等边 ABD使点CD在AB的同侧再以CD为一边作等边 ECD使点CE在 sketch template

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