已知cd垂ab于d,ef垂ab于e,角1等角2,角3等75度,求角acb的度

来源：学生作业帮编辑：拍题作业网作业帮分类：英语作业时间：2024/04/28 01:00:40

The Nike Air Flightposite II LE is set to release this Saturday at Foot Locker and its House of Hoops locations making it a fairly easy shoe to get a hold of.
This is the 10th anniversary of the shoe that Kevin Garnett made famous and while the past has seen several color ways released it will be dropping in a triple black color way.Anybody else juiced about this one?Who’s copping?Via Footlockerunlocked
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已知cd垂ab于d,ef垂ab于e,角1等角2,角3等75度,求角acb的度如图,AB//CD,EF垂直于AB于E,EF交CD于F,已知角1-60度,则角2( ) 已知梯形ABCD中AB平行于CD,E、F分别是AB、CD的中点,且角D+角C=90度,试证明EF=1/2（CD—AB）如图,已知:在三角形ABC中,角ACB=90度,CD垂直于AB于D,EF垂直于AB于F,且EC=EF求证:FG平行于AC 已知三角形ABC中,AB等于AC,CD平分角ACB,CE垂直AB于E,角DCE等于57度,求角ACB的度数已知AB∥CD,直线EF分别交AB、CD于点E、F,EG平分∠BEF交CD于点G,如果角1=108度,求角2的度数. 如图在Rt三角形ABC中角ACB=90度,CD垂直AB于D,BE平分角ABC交AC于E,交CD于H,EF垂直AB于F,连如图,已知CD垂直AB于点D,E是BC上一点,EF垂直AB于点F,角1=角2求证角AGD=角acb 如图AB平行于CD直线EF分别交AB,CD于点E,F,EG平分角AEF,交CD于G.已知角1等于40度,求角2 已知在直角三角形ABC,角ACB=90度,EF垂直平分线,交AC、AB于点E、F,EF=EC,求角A的度数. 已知在三角形ABC中,角ACB等于90度,CD垂直AB于D若角B等于 30度,CD等于6,求AB的长已知梯形ABCD,AB平行于CD,EF分别是AB ,CD的中点,且角D加角C等于90度,证明EF=1/2(CD-AB)