25．如图在 abc中 De分别是acab上的点 Bd与ce交于点o 给出下列三个条件�∠ebo＝∠dco�∠beo＝∠cdo Sketch Coloring Page

25．如图在 abc中 De分别是acab上的点 Bd与ce交于点o 给出下列三个条件�∠ebo＝∠dco�∠beo＝∠cdo Template

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More Sketches

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

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sketch #3699 Dragon by Ana Cecilia

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tomski25．如图在 abc中 de分别是acab上的点 coloring

25．如图在 ABC中 DE分别是ACAB上的点 BD与CE交于点O 给出下列三个条件�∠EBO＝∠DCO�∠BEO＝∠CDO sketch template

25．如图在 ABC中 DE分别是ACAB上的点 BD与CE交于点O 给出下列三个条件�∠EBO＝∠DCO�∠BEO＝∠CDO 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

题目如图在 ABC中边ABAC的垂直平分线分别交BC于DE．1若BC=10求 ADE的�长 2 设直线DMEN交于点 sketch template

百度教育 sketch template

如图 ABC中BDCE是 ABC的两条高点FM分别是DEBC的中点．求证FM⊥DE．题目和参考答案——青夏教育精英家教�—— sketch template

如图 ABC中 DE分别是ACAB上的点 BD与CE交于点O 给出下列三个条件 �∠EBO＝∠DCO�∠BEO＝∠CDO sketch template

如图所示在等边 ABC中点DE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

25．如图在 ABC中 DE分别是ACAB上的点 BD与CE交于点O 给出下列三个条件�∠EBO＝∠DCO�∠BEO＝∠CDO sketch template

如图在 ABC中DE分别是边ABAC的中点BC=8则DE=．题目和参考答案——青夏教育精英家教�—— sketch template

如图 ABC中DE分别是ABAC上的点DE∥BCDE＝1BC＝3AB＝6则AD的长为 A．1 B．15 C．2 D sketch template

如图在 ABC中分别以ABAC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，在 ABC中，AB=4，AC=3，BC=5，DE是BC的垂直平分线，DE分别交BC、AB于点D、E．1求证： ABC为直角三角形． 2求 sketch template

如图 ABC中DE分别是ABAC上的点DE∥BCDE＝1BC＝3AB＝6则AD的长为 A．1 B．15 C．2 D sketch template

如图�，已知 ABC的内切圆⊙O与AB、BC、AC分别相切于点D、E、F，且EF=DE（1）判断 ABC的形状，并证明你的结论；（2 sketch template

25．如图在 ABC中 DE分别是ACAB上的点 BD与CE交于点O 给出下列三个条件�∠EBO＝∠DCO�∠BEO＝∠CDO sketch template

如图，在 ABC中，AD、BE分别是边BC、AC上的中线，AD与BE交于点O，点F、G分别是BO、AO的中点，联结DE、EG、GF、FD．（1 sketch template

如图在等要直角三角形ABC中O是斜边AC的中点P是斜边AC上的一个动点D为BC上的一点且PB=PD垂足为点E 1求证PE sketch template

1证法回顾证明三角形中位线定理．已知如图1DE是 ABC的中位线．求证．证明添加辅助线如图1在 ABC中延长DE sketch template

如图在 ABC中点DE分别在边ACAB上BD与CE交于点O给出下列四个条件�∠EBO=∠DCO�BE=CD�OB=OC sketch template

15．如图在 ABC中∠C=90°AB=10经过点C且与边AB相切的动圆与CACB分别交于点DE则线段DE长度的最小值是 sketch template

已知如图在等边三角形ABC中DE分别为BCAC上的点且AE＝CD连结ADBE交于点P作BQ⊥AD垂足为Q．求证BP sketch template

如图 ABC中点D在AC上E在AB上且AB=ACBC=CDAD=DE=BE 1求证 BCE≌ DCE 2求∠EDC的度数 sketch template

26．如图已知 ABC中DE分别是ACAB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

如图以 ABC的ABAC为一边�外作正方形ABGF和ACDE两个正方形对角线交点分别是MNP是BC中点求证PM＝PN．题目和 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，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Rt ABC 中∠C=90°AC=BCAB=8点D为AB的中点若直角MDN绕点D旋转分别交AC于点E交BC于点M则下列 sketch template

26．如图已知 ABC中DE分别是ACAB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

如图1在 ABC中AC=BC∠ACB=90°点D为AB边中点以点D为顶点作∠PDQ=90°DPDQ分别交直线ACBC于EF sketch template

如图在ABC中AB=ACBD和CD分别是ABC和ACB的平分线EF过D点且FF∥BC图中等腰三角形共有 A2个 B3个 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

图，三角形ABC中，EF分别在AB，AC上，DE垂直于DF，D是边上BC的中点，则BE CF答案及详细解析过程——青夏教育精英家教�—— sketch template

如图所示在 ABC中AB=AC∠A=36°BDCE分别为∠ABC与∠ACB的角平分线且相交于点F则图中的等腰三角形有 A．6个 B sketch template

如图AD是 ABC的角平分线DE⊥AB于点ES ABC=9DE=2AB=5则边AC的长是 A3B4C5D6 题目和 sketch template

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