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 Coloring Page

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 Template

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anonymous

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tomski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 coloring

如图，在Rt ABC中，∠ C=90°，点D在AB上，以AD为直径的⊙ O与BC相交于点E，与AC相交于点F，AE平分∠ BAC1求证 sketch template

如图Rt ABC中∠ABC=90°以AB为直径的⊙O交AC于点DE为BC的中点连接DEOD．1求证DE是⊙O的切线2 sketch template

已知如图在Rt ABC中∠C=90°以AC为直径作⊙O交AB于D过O作OE∥AB交BC于E．1求证ED是⊙O的切线．2 sketch template

如图在Rt ABC中∠C=90°BD是∠ABC的平分线交AC于D若CD=mAB=n则 ABD的面积是 A．mnB．12mnC sketch template

题目如图等腰Rt ABC中∠ABC＝90°AB＝BC点AB分别在坐标轴上．1如图1若点C的横坐标为5直接写出点B的坐标 sketch template

如图Rt ABC纸片中∠C=90°AC=6BC=8点D在边BC 上以AD为折痕将 ABD折叠得到 AB′DAB′与边BC交于点E sketch template

如图在Rt ABC中∠ACB=90°以AC为直径作⊙O交AB于点DE为BC的中点连接DE并延长交AC的延长线于点F．1求证DE sketch template

如图在Rt ABC中∠ACB=90°AC=4BC=3D是边AB上一点联结CD把 ACD沿CD所在的直线翻折点A落在点E的位置 sketch template

14．如图Rt ABC中AC=8BC=6∠C=90°⊙I分别切ACBCAB于DEF求Rt ABC的内心I与外心O之间的距离 sketch template

如图在Rt ABC中∠ACB=90°以AC为直径的⊙O与AB边交于点D过点D作⊙O的切线交BC于点E．1求证点E是边BC的中点 sketch template

如图在Rt ABC中∠C=90°BD是∠ABC的平分线交AC于D若CD=mAB=n则 ABD的面积是 A．mnB．12mnC sketch template

如图Rt ABC中∠C=90°O为AB上一点以O为圆心OB长为半径的圆交BC边于点D与AC边相切于点E．1求证BE平分∠ sketch template

如图Rt ABC中∠ACB=90°AC=3BC=1将Rt ABC绕C点旋转90°后为Rt A′B′C′再将Rt A′B′C′绕B点 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=6BC=8点F在边AC上并且CF=2点E为边BC上的动点将 CEF沿直线EF翻折点C落 sketch template

15．如图Rt ABC中∠C=90°∠ABC=30°AB=6．点D在AB边上点E是BC边上一点不与点BC重合且DA=DE则 sketch template

问题如图1在Rt ABC中∠C=90°⊙O是Rt ABC的内切圆切点分别是DEF若三角形三边长分别记为BC=aAC=bAB sketch template

4．如图在Rt ABC中∠C=90°． 1若AC=5BC=12则AB= TanA= ∠A≈ 2若AC=3AB=5则 sketch template

如图Rt ABC中∠C=90°⊙O为 ABC的内切圆若AC=6BC=8求⊙O半径．题目和参考答案——青夏教育精英家教�—— sketch template

如图在Rt ABC中∠C=90°AC=6∠A的平分线交BC于点DDE⊥AB于点E且BE=4求DE的长．题目和参考答案——青夏 sketch template

如图6在Rt ABC中 ∠C＝90°BD是∠ABC的平分线交AC于D若CD＝4AB＝8则 ABD的面积是 sketch template

如图在Rt ABC 中∠C=90°AC=BCAB=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

如图Rt ABC中∠ABC＝90°以AB为直径的⊙O交AC于点DE是BC的中点连接DEOE．1判断DE与⊙O的位置关系并说明 sketch template

如图在Rt ABC中∠ACB=90°以AC为直径的⊙O与AB边交于点D过点D作⊙O的切线交BC于点E．1求证点E是边BC的中点 sketch template

如图在 Rt ABC 中 ∠C = 90 °点 D 大学搜题酱 sketch template

如图Rt ABC∠C=90°AD平分∠CABDE⊥AB于E则下列结论中不正确的是 A．BD ED=BCB．DE平分∠ADBC sketch template

已知Rt ABC中∠ACB=90°AC=6BC=8 Word文档免费下载文档大全 sketch template

如图Rt ABC中∠ACB=90°AC=3BC=1将Rt ABC绕C点旋转90°后为Rt A′B′C′再将Rt A′B′C′绕B点 sketch template

如图在Rt ABC中∠C=90°以点C为圆心作⊙C与AB切于点D过点AB分别作⊙C的切线AFBE切点为FE点．求证AF∥ sketch template

题目在Rt ABC中∠C=90°BC=3AC=4点P在以C为圆心5为半径的圆上连结PAPB．若PB=4则PA的长为 sketch template

如图Rt ABC中∠ABC=90°BA=BC=2分别过AC作平面ABC的垂线AA′和CC′AA′=h1CC′=h2且h1＞h2 sketch template

如图在Rt ABC 中∠C=90°AC=BCAB=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

如图在 Rt ABC 中 ∠C = 90 °点 D 大学搜题酱 sketch template

如图在Rt ABC中∠BAC=90°∠C=30°以边AC上一点O为圆心OA为半径作圆恰好经过边BC的中点D并与边AC相交于另一点 sketch template

如图所示在Rt ABC中∠ABC＝90°．将Rt ABC绕点C顺时针方�旋转60°得到 DEC点E在AC上再将Rt ABC沿着AB所在 sketch template

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