(2014?牡丹江一模)如图,在△ABC中,AB=AC,AD、BE是△ABC的两条高,连接DE交AB于点O,则下列结论中,

2025-04-27 11:05:11
推荐回答(1个)
回答1:

①在△ABC中,AB=AC,AD是△ABC的高,
∴BD=DC,
∵BE⊥AC,
∴ED=

1
2
BC.
故①正确;
②∵ED=
1
2
BC=BD=DC,
∴∠BAC=∠EDC,∠EBD=∠BED,
∴∠BAC=2∠EBD=2∠BED;
故②正确;
③∵∠BAC=∠EDC,
∴∠ABE=∠EDA,
∴△AOD∽△BOE.
故③正确;
④BC2=BE2+EC2
=BE2+(EA+AC)2
=BE2+EA2+AC2+2EA?AC
=AB2+AC2+2EA?AC
=2AC2+2EA?AC
=2AC(AC+EA)
=2CA?CE.
故④正确;
⑤图中相似的三角形有5对.
故⑤错误.
故选:D.