如图ac⊥bccd⊥abde⊥bc分别交bcabbc于cde下列说法中不正确的是aac是abc的高 Bde是bcd的高 Sketch Coloring Page

如图ac⊥bccd⊥abde⊥bc分别交bcabbc于cde下列说法中不正确的是aac是abc的高 Bde是bcd的高 Template

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يآآسر محمد
sketch #2818 Hellboy by يآآسر محمد

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sketch #5073 Broken Heart by Islam Mourad

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tomski如图ac⊥bccd⊥abde⊥bc分别交bcabbc于cde下列说法中不正确的是aac是abc的高 bde是bcd的高 coloring

如图四边形ABCD内接于⊙OAB是⊙O的直径AC和BD相交于点E且1求证BC=CD2分别延长ABDC交于点P过点A作 sketch template

题目如图已知AC⊥BCAD⊥BDE为AB的中点1如图1求证 ECD是等腰三角形2如图2CD与AB交点为F若AD sketch template

如图AB是⊙O的直径点C为弧BD的中点CF为⊙O的弦且CF⊥AB垂足为E连接BD交CF于点G连接CDADBF1求证 sketch template

如图AC⊥BCCD⊥ABDE⊥BC分别交BCABBC于CDE下列说法中不正确的是AAC是ABC的高 BDE是BCD的高 sketch template

三角形的有关概念提高巩固练习 Word文档在线阅读与下载无忧文档 sketch template

如图1已知AB是⊙O的直径弦CD⊥AB于点E点M在⊙O上∠M=∠D1判断BC与MD的位置关系并说明理由2若AE=8BE sketch template

如图在 ABC中∠C=90°AC=BCAD是角平分线DE⊥AB．垂足为E．求证BE=DE=CD．题目和参考答案——青夏教育精英家教�—— sketch template

如图 ABC内接于⊙OAB=AC过点C作⊙O的直径CD连接BD．1求证∠BDC=2∠ABD2连接OA求证OA∥BD的 sketch template

题目如图直线y＝x 6与x轴y轴交于AB两点点C在第四象限BC⊥AB且BC＝AB1如图1求点C的坐标2如图2D sketch template

如图在Rt ABC中AC=1BC=xD为斜边AB的中点．将 BCD沿直线CD翻折．若在翻折过程中存在某个位置使得CB⊥AD则x的 sketch template

如图，以 ABC的边AB、AC为边�外作正方形ABDE和正方形ACFG，AH⊥BC，垂足为点H，交EG于点M．求证：EM=MG．百度教育 sketch template

如图⊙O的两条弦ABCD互相垂直垂足为E且AB=CD．1求证AC=BD．2若OF⊥CD于FOG⊥AB于G问四边形 sketch template

如图AC⊥BCD在AB上∠CDA=90°则线段CDCD的长度是点C到直线AB的距离线段BC的长度是点B到直线ACAC的距离．题目和 sketch template

如图D为直角 ABC的斜边AB上一点DE⊥AB交AC于E如果 AED沿DE翻折A恰好与B重合联结CD交BE于F如果AC=8 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所示，已知AB为圆O的直径，CD是弦，且AB⊥CD于点E，连接AC、OC、BC． sketch template

$已知线段ABCD为⊙O的两条弦且AB⊥CD于点H连接ACBCBD．1如图1过圆心O作OE⊥BD于点E求证OE=$frac sketch template$

如图在 ABC中∠C=90°AC=BC点D为AB的中点点EF分别为ACBC边上的动点且CE=BF当点EF分别在边ACBC sketch template

如图在Rt ABC中∠C=90°DE⊥ABAC=BE=15BC=20则四边形ACED的面积为 75 96——青夏教育精英家教�—— 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在梯形ABCD中AD∥BCAB=AC∠BAC=90°BC=BDAC与BD相交于O．求证CD=CO．题目和参考答案——青夏 sketch template

如图AB为⊙O的直径点C是⊙O上的一点CD⊥AB垂足为点DCF⊥AF且CF=CDAF交⊙O于点EBE交AC于点M．1求证 sketch template

如图1在四边形ABCD中如果对角线AC和BD相交并且相等那么�们把这样的四边形称为等角线四边形．1�在“平行四边形矩形菱形 中 sketch template

如图 ABC内接于⊙OAB=AC连接AO并延长交BC于点D1求证BD=CD2如图点P为弧AB上一点连接BPCP作AH sketch template

12．如图AB是⊙O的弦OC⊥AB于点D交⊙O于点C若⊙O的半径为10CD＝4那么AB的长为．——青夏教育精英家教�—— sketch template

如图在 ABC中AC=BC∠C=90°AD是 ABC的角平分线DE⊥AB垂足为E．1若CD=5求AC的长．2求证AB sketch template

如图在 ABC中以AB为直径的⊙O交BC于点D且BD=CD交直线AC于点E连接BE．1如图1求证∠CAB=2∠CBE2 sketch template

如图在四面体ABCD中AB⊥面BCD面ABC⊥面ACD且∠ACB=∠CBD=45°1求证BC⊥CD2求直线AC与平面 sketch template

如图在 ABC中AB=ACAD⊥BC于点DDE⊥AB于点EDF⊥AC于点F下列结论�∠BAD=∠CAD�AD上任意一点到AB sketch template

如图在 ABC中AD平分∠BACDG⊥BC且平分BC于点GDE⊥AB于EDF⊥AC于F．1证明BE=CF2如果AB=12 sketch template

题目如图在 ABC中∠C=90°BC=ACD是AC上一点AE⊥BD交BD的延长线于EAE=BD且DF⊥AB于F求证CD sketch template

已知四边形ABCD内接于⊙O对角线ACBD交于点P．1如图1设⊙O的半径是r若AB CD=πr求证AC⊥BD2如图2过 sketch template

题目问题背景如图1等腰 ABC中AB=AC∠BAC=120°作AD⊥BC于点D则D为BC的中点∠BAD=∠BAC=60 sketch template

探究1如图1在ABC与ADE中AB=ACAD=AE∠BAC=∠DAE=90°连结BDCE．请写出图1中所有全等的三角形 sketch template

如图所示在四边形ABCD中AD∥BCE为CD的中点连接AEBEAB=AD BC．求证BE⊥AE．题目和参考答案——青夏教育精英 sketch template

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