三角形abc中，d、e分别是ab边ac边上的点，且ad=ae，be=cd，求证ab=ac 知乎 Sketch Coloring Page

三角形abc中，d、e分别是ab边ac边上的点，且ad=ae，be=cd，求证ab=ac 知乎 Template

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tomski三角形abc中，d、e分别是ab边ac边上的点，且ad=ae，be=cd，求证ab=ac 知乎 coloring

三角形中线的定理？ 知乎 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

这个几何题怎么解决？ 知乎 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

25．如图在 ABC中 DE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 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

百度教育 sketch template

如图在等边 ABC中，D、E分别是BC、AC上的点，且AE=CD，AD与BE相交于F，CF⊥BE．求证：（1）BE=AD；（2）BF=2AF 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

已知在等边三角形ABC中DE分别为ABAC上的点且BD=AEEB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

百度教育 sketch template

如图 ABC中 DE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 template

百度教育 sketch template

14．如图点DEF分别是 ABC三边上的中点．若 ABC 的面积为12则 DEF的面积为．——青夏教育精英家教�—— sketch template

百度教育 sketch template

三角形“四心”有关性质及证明（内心篇） 知乎 sketch template

三角形ABC中，D、E分别是AB边AC边上的点，且AD=AE，BE=CD，求证AB=AC 知乎 sketch template

如图 ABC∽ A′BC′ADA′D′分别是这两个三角形的高EFE′F′分别是这两个三角形的中位线ADA′D′与EFE′F′相等吗 sketch template

如图已知 ABC中AB=ACM是BC的中点DE分别是ABAC边上的点且BD=CE1求证MD=ME．2若D为AB的 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

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

已知如图在等边三角形ABC中MN分别是ABAC的中点D是MN上任意一点CDBD的延长线分别与ABAC交于FE若1CE sketch template

25．如图在 ABC中 DE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 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

题目1如图1在四边形ABCD中FE分别是BCAD的中点连结EF并延长分别与BACD的延长线交于点MN则∠BME=∠ sketch template

百度教育 sketch template

如图在正三角形ABC中DEF分别为各边的中点GHIJ分别为AFADBEDE的中点．将 ABC沿DEEFDF折成三棱锥 sketch template

图，三角形ABC中，EF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

百度教育 sketch template

如图所示，已知边长为4的正三角形ABC中，E，F分别为BC和AC的中点，PA=2，且PA⊥平面ABC，设Q是CE的中点（1）求证：AE∥平面 sketch template

百度教育 sketch template

在三角形ABC中∠B＝60°BA＝24cmBC＝16cm现有动点P从点A出�沿射线AB�点B方�运动动点Q从点C出�沿射线CB sketch template

如图等边三角形ABC中DE分别为ABBC边上的两个动点且总使BD=CEAE与CD交于点FAG⊥CD于点G则以下结论∠AFG sketch template

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