已知AB//CD,BD平分∠ABC交AC于O,CE平分∠DCG,若∠ACE=90°,求证BD⊥AC如题G与B.C共线 CE平分∠DCG

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已知AB//CD,BD平分∠ABC交AC于O,CE平分∠DCG,若∠ACE=90°,求证BD⊥AC如题G与B.C共线CE平分∠DCG已知AB//CD,BD平分∠ABC交AC于O,CE平分∠DCG,若∠

已知AB//CD,BD平分∠ABC交AC于O,CE平分∠DCG,若∠ACE=90°,求证BD⊥AC如题G与B.C共线 CE平分∠DCG
已知AB//CD,BD平分∠ABC交AC于O,CE平分∠DCG,若∠ACE=90°,求证BD⊥AC
如题
G与B.C共线 CE平分∠DCG

已知AB//CD,BD平分∠ABC交AC于O,CE平分∠DCG,若∠ACE=90°,求证BD⊥AC如题G与B.C共线 CE平分∠DCG
你这个问题少条件啊,E点G点在哪儿啊?
因为G在BC线上（应该在延长线上）,因为AB//CD,所以

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已知AB//CD,BD平分∠ABC交AC于O,CE平分∠DCG,若∠ACE=90°,求证BD⊥AC 在一小时之内已知AB//CD,BD平分∠ABC交AC于O,CE平分∠DCG,若∠ACE=90°,求证BD⊥AC如题G与B.C共线 CE平分∠DCG 如图:已知AB∥CD,BD平分∠ABC交AC与O,CE平分∠DCG,∠ACE=90°,那么BD和AC是否垂直?请说明理由. 已知△ABC中,BD平分∠EBC,CD平分∠BCF交BD于D. 几何角平分线如图,已知BD平分∠ABC,CD平分∠ACE交BD于点D,求证：AB平分∠CAF. 已知三角形ABC中,AB=2AC,AD平分∠BAC,且AD=BD,求证：CD⊥AC. 已知,如图,AB=AC,∠A=108°,BD平分∠ABC交AC于D,求证：BC=AB+CD点E已经画出来了，可是没有说 CD=CE啊、如图所示已知AB平行CD,BD平分角ABC交AC于O,CE平分角DCG.若角ACE90°,请判断如图所示已知AB平行CD,BD平分角ABC交AC于O,CE平分角DCG.若角ACE90°,请判断BD与AC的位置关系,并说明理由如图,△ABC是等腰三角形,AB=AC,∠A=108,BD平分∠ABC,交AC于点D.求证BC=AB+CD 如图,在△ABC中,∠BAC=108°,AB=AC,BD平分∠ABC,交AC于D,求证：BC=CD+AB . 在△ABC中,AB=AC,∠A=108°,BD平分∠ABC,交AC于D点.求证：BC=AB+CD 在△ABC中,AB=AC,∠A=108°,BD平分∠ABC,交AC于D点.求证：BC=AB+CD 已知：在三角形ABC中,AB=AC,BD平分∠ABC,CD平分∠ACB,过点D作 EF∥BC,分别已知：在三角形ABC中,AB=AC,BD平分∠ABC,CD平分∠ACB,过点D作 EF∥BC,分别交AB/AC于E/F两点（如图1） ①图中共有多少个等腰三角已知△ABC内接于圆O,AD平分∠BAC交圆O于D交BC于E（AD不为直径）,连BD和CD,证明：AB×AC+BD×DC=AD² 已知：如图,在RT△ABC中,∠C=90°,BD平分∠ABC,BD交AC于点D,DE⊥AB,且AD=2CD.求证；∠A=30° 已知:如图Rt三角形ABC中,∠C＝90°,BD平分∠ABC,BD交AC于点D,DE⊥AB,且AD=2CD.求证:∠A=30° 步骤详细. 如图,△abc中,∠abc中,∠bac=108,ab=ac,bd平分∠abc,交ac于d,求证bc=cd+ab 两种方法如图,△abc中,∠abc中,∠bac=108,ab=ac,bd平分∠abc,交ac于d,求证bc=cd+ab 两种方法如图,AB=AC,∠A=108º,BD平分∠ABC交AC于D,求证：BC=AB+CD