如图已知d是 Abc的bc边的中点de⊥acdf⊥ab垂足分别为ef且df=de则 Abc是等腰三角形吗请说明理由．——青夏 Sketch Coloring Page

如图已知d是 Abc的bc边的中点de⊥acdf⊥ab垂足分别为ef且df=de则 Abc是等腰三角形吗请说明理由．——青夏 Template

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anonymous

anonymous
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anonymous

anonymous
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sketch #90 JF Style!

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sketch #3398 Sharingan by حيرتني بصمتها

anonymous

anonymous
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tomski如图已知d是 abc的bc边的中点de⊥acdf⊥ab垂足分别为ef且df=de则 abc是等腰三角形吗请说明理由．——青夏 coloring

如图�，已知 ABC是等腰直角三角形，∠BAC＝90°，点D是BC的中点．作正方形DEFG，使点A、C分别在DG和DE上，连接AE、BG．（1 sketch template

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

16．如图已知D E分别是 ABC的边BC和边AC的中点连接DEAD若S=24cm求 DEC的面积——青夏教育精英家教�—— 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D是 ABC的BC边的中点DE⊥ACDF⊥AB垂足分别为EF且DF=DE则 ABC是等腰三角形吗请说明理由．——青夏 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已知等边 ABC以边BC为直径的圆与ABAC分别交于点D点E．过点D作DF⊥AC垂足为F．1判断DF与⊙O的位置关系并证明 sketch template

如图�，已知 ABC是等腰直角三角形，∠BAC＝90°，点D是BC的中点．作正方形DEFG，使点A、C分别在DG和DE上，连接AE、BG．（1 sketch template

如图已知点D是AB边的中点AF∥BCCGGA=31BC=8则AF= ．题目和参考答案——青夏教育精英家教�—— 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ACBC于EF 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，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D是等边 ABC的边AB上一点E是BC延长线上一点CE=DA连接DE交AC于F过D点作DG⊥AC于G点．证明下列结论1AG sketch template

如图AB是半圆⊙O的直径 ABC的两边ACBC分别交半圆于DE且E为BC的中点已知∠BAC=50°则∠C= A 55 sketch template

如图 ABC为等边三角形以边BC为直径的半圆与边ABAC分别交于DF两点过点D作DE⊥AC垂足为点E．1判断DE与⊙O的位置 sketch template

如图在 ABC中AB=BCBE平分∠ABCCD⊥AB于点DCD=BD点H是BC边的中点连接DH交BE于点G连接CG．1 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∠A=90°AB=ACD 是BC的中点点EF分别是ABAC上的点并且BE=AF1猜想并写出AD与BC所满足 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DE分别为 ABC的边ABAC上的点BE与CD相交于O点现有四个条件�AB＝AC�OB＝OC�∠ABE＝∠ACD sketch template

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

如图在等腰Rt ABC中∠C=90°AC=BC=8F是AB边上的中点点DE分别在ACBC边上运动且保持AD=CE连接DE sketch template

21如图，已知等腰直角三角形ABC，点P是斜边BC上一点不与B，C重合，PE是 ABP的外接圆⊙O的直径．1求证： APE是等腰直角 sketch template

如图在Rt ABC中∠ABC=90°BA=BC点D是AB的中点连接CD过点B作BG⊥CD分别交CDCA于点EF与过点A且 sketch template

如图 ABC中点D在线段BC上且 ABC∽ DBA则下列结论一定正确的是 AAB2=BC·BD BAB2=AC·BD CAB sketch template

题目1如图1在四边形ABCD中FE分别是BCAD的中点连结EF并延长分别与BACD的延长线交于点MN则∠BME=∠ 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已知 ABC中AB=ACM是BC的中点DE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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

如图，AB是半圆O的直径，C是半圆O上一点（不与点A、B重合），D是AC的中点，DE⊥ AB于点E，过点C作半圆O的切线，交ED的延长线于 sketch template

如图已知C是AB的中点D是AC的中点E是BC的中点 1若AB=18cm求DE的长2若CE=5cm求DB的长——青夏 sketch template

如图以 ABC的边AB为直径作⊙O交BC于D过D作⊙O的切线交AC于E要使得DE⊥AC则 ABC的边必须满足的条件是 ．题目和参考 sketch template

31． 如图已知D是BC的中点BE⊥AE于ECF⊥AE于F 求证BE = CF——青夏教育精英家教�—— sketch template

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