PM⊥BD于M,PN⊥AD于N,BD=AD,PM=PN.求证:OB=OA

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PM⊥BD于M,PN⊥AD于N,BD=AD,PM=PN.求证:OB=OA
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PM⊥BD于M,PN⊥AD于N,BD=AD,PM=PN.求证:OB=OA
PM⊥BD于M,PN⊥AD于N,BD=AD,PM=PN.求证:OB=OA

PM⊥BD于M,PN⊥AD于N,BD=AD,PM=PN.求证:OB=OA
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PM⊥BD于M,PN⊥AD于N,BD=AD,PM=PN.求证:OB=OA 如图,OD平分∠AOB,在OA、OB边上取OA=OB,PM⊥BD,于M,PN⊥AD于N,求证PM=PN. 如图,BD是∠ABC的平分线,AB=BC,P在BD上,PM⊥AD于M,PN⊥CD于N求证PM=PN 如图,已知BD是∠ABC的平分线,AB=BC.点D在射线BD上,PM⊥AD于M,PN⊥CD于N.求证PM=PN.拜托有谁做过这题, 已知:如图,BD为∠ABC的平行线,AB=BC,点P在BD上,PM⊥AD于M,PN⊥CD于N.求证:PM=PN. 已知:如图,BD为∠ABC的平行线,AB=BC,点P在BD上,PM⊥AD于M,PN⊥CD于N.求证:PM=PN. 已知:如图所示,BD为∠ABC的平分线,AB=BC,点P在BD上,PM⊥AD于M,PN⊥CD于N,判断PM与PN的关系 已知:如图所示,BD为∠ABC的平分线,AB=BC,点P在BD上,PM⊥AD于M,PN⊥CD于N,判断PM与PN的关系 如图,BD是∠ABC的平分线,AB=BC 点P在BD上.PM⊥AD于M,PN垂直CD于N,证明:PM=PN OK的话悬赏×3!我是新手,初学者能看得懂的那种!四边形ABCD中,对角线AC,BD交于点O,AE⊥BD于E,在AD上任取一点P作PM⊥AC于M,PN⊥BD于N,若AE=a,求PM+PN的长! 如图,BD是∠ABC的角平分线,AB=BC,点P在BD上,PM⊥AD与点M,PN⊥CD于点N,求证:PM=PN. 如图,BD是∠ABC的角平分线,AB=BC,点P在BD上,PM⊥AD与点M,PN⊥CD于点N,求证:PM=PN. 已知,如图BD为∠ABC的平分线,AB=BC,点P在BD上,PM⊥AD于M,PN⊥CD于N,求证PM=PN. 在四边形ABCD中,BD为∠ABC的平分线,AB=BC,点P在BF上,PM⊥AD于M,PN⊥CD于N,求证:PM=PM 已知正方形abcd的边长a,E是对角线BD上一点,BE是a,P是EC上任意一点,PM⊥BD于M,PN⊥BC于N,求PM+PN 如图所示,BD平分∠ABC,AB=BC,点P在BD上,PM垂直AD,PN⊥CD,M、N为垂足.求证:PM=PN OD平分∠AOB,OA=OB,P是OD上的一点,PM⊥BD于M,PN⊥AD于N,求证:PM=PN(用角平分线的意义来求) 梯形ABCD中,P,M,N分别为AD,AB,CD上的点,且PM平行于BD,PN平行于AC求证 PM/BD+PN/AC=1