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数学
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已知非Rt△ABC中,∠A=45°,高BD、CE所在的直线交于点H,画出图形并求出∠BHC的度数.
人气:470 ℃ 时间:2019-09-29 06:14:17
解答
①如图1,△ABC是锐角三角形时,
∵BD、CE是△ABC的高线,
∴∠ADB=90°,∠BEC=90°,
在△ABD中,∵∠A=45°,
∴∠ABD=90°-45°=45°,
∴∠BHC=∠ABD+∠BEC=45°+90°=135°;
②△ABC是钝角三角形时,∵BD、CE是△ABC的高线,
∴∠A+∠ACE=90°,∠BHC+∠HCD=90°,
∵∠ACE=∠HCD(对顶角相等),
∴∠BHC=∠A=45°,
综上所述,∠BHC的度数是135°或45.
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