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oae与cd交于点gac与bd交于点f Sketch Coloring Page

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oae与cd交于点gac与bd交于点f Template

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tomski10．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oae与cd交于点gac与bd交于点f coloring

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OAE与CD交于点GAC与BD交于点F 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OAE与CD交于点GAC与BD交于点F 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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在等边 ABC中点DE分别在边BCAB上且BD=AEAD与CE交于点F则∠DFC的度数为 A．60°B．45°C sketch template

1如图�所示已知C是线段AB上一点分别以ACBC为边长在AB的同侧作等边 ADC和 CBE连接AE和DB证明AE=DB2 sketch template

如图所示已知 ABC和 ECD都是等腰直角三角形∠ACB=∠DCE=90．D为AB边上 点若AB=10cmAD=3cm求AE的长 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

如图所示已知点D在 ABC的边AB上连结CD∠1=∠BAD=4AC=5 求 BD的长——青夏教育精英家教�—— sketch template

如图1所示已知 ABC中∠BAC=90°AB=ACAE是过点A的一条直线且B点和C点在AE的同侧BD⊥AE于D点CE⊥AE于E点 sketch template

如图已知 ABC ADE均为等边三角形点D是BC延长线上一点连结CE求证1∠BAD=∠CAE 2 BAD≌ CAE3 sketch template

如图所示已知在 ABC中∠BAC＝70°AC=6AD是 ABC的角平分线则∠BAD= BE是 ABC的中线则AE=CE= CF sketch template

如图所示已知梯形ABCD中AB=2CD点E分有�线段AC所成的比为λ双曲线过CDE三点且以AB为焦点．当时求双曲线的 sketch template

如图所示，已知三角形ABC为等腰直角三角形，且EC垂直AC于C，AE=BF，判断AE和BF的位置关系。百度知道 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

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

如图所示已知边长为3的等边 ABC点F在边BC上CF=1点E是射线BA上一动点以线段EF为边�右侧作等边 EFG直线EGFG交 sketch template

如图所示在 ABC中∠ABC=90°AB=8BC=6若DE是 ABC的中位线延长DE交 ABC的外角∠ACM的平分线于点F求线段 sketch template

15．如图所示已知等腰直角三角形ABC中∠ACB=90°直线L经过点C AD ⊥LBE⊥L垂足分别为DE． 1证明 ACD sketch template

百度教育 sketch template

21．如图所示已知D是等腰三角形底边BC上一点它到两腰的距离分别为请你指出当D点在什么位置时并加以证明——青夏教育精英家教�—— sketch template

如图所示已知 ABC的�长是21OBOC分别平分∠ABC和∠ACBOD⊥BC于D且OD=3则 ABC的面积是多� 题目和参考答案 sketch template

10．如图中BD平分∠ABCDE∥BC交AB于点EEF∥AC交BC于F试说明BE=CF——青夏教育精英家教�—— sketch template

6．．如图所示∠1∠2∠3的大小关系是 ． A．∠1>∠2>∠3 B．∠3>∠1>∠2 C．∠2>∠1>∠3 D．∠3>∠2>∠1 sketch template

如图O是等边 ABC内的一点已知∠AOB=110°∠COD=60°∠BOC=α BOC≌ ADC．1求证 COD是等边三角形 sketch template

如图1所示已知AB//CDAD和BC相交于点O若则 图1 图2 图3——青夏教育精英家教�—— sketch template

如图所示已知点AEFD在同一条直线上AE=DFBF⊥ADCE⊥AD垂足分别为FEBF=CE求证AB∥CD．题目和参考 sketch template

如图所示已知DE分别是 ABC的边BCAB上的点ADCE交FBFDE交于G过G作BC的平行线MN交ABCEAC于MH sketch template

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

如图1 已知等腰三角形ABC 则图2所示的四个图中可能是 ABC的直观图的是 1 2 ——青夏教育精英家教�—— sketch template

如图已知DE∥BCBD是∠ABC的平分线∠EDC＝109° ∠ABC＝50°则∠A 度∠BDC＝度——青夏教育精英家教�—— sketch template

如图所示已知在 ABC中∠BAC＝70°AC=6AD是 ABC的角平分线则∠BAD= BE是 ABC的中线则AE=CE= CF sketch template

如图所示已知 ABC中AB=AC=10cmBC=8cm点D为AB的中点．如果点P在线段BC上由B出��C点运动同时点Q在线段CA上由 sketch template

6．如图已知ΔABC和ΔABD都是⊙O的内接三角形AC和BD相交于点E则与ΔADE相似的三角形是 A．ΔBCE B．ΔABC C sketch template

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