如图,Rt△AB′C′是由Rt△ABC绕点A顺时针旋转得到的,连接CC′交斜边于点E,CC′的延长线交BB′于点F.(1)若AC=3,AB=4,求CC′:BB′
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![如图,Rt△AB′C′是由Rt△ABC绕点A顺时针旋转得到的,连接CC′交斜边于点E,CC′的延长线交BB′于点F.(1)若AC=3,AB=4,求CC′:BB′](/uploads/image/z/1038318-6-8.jpg?t=%E5%A6%82%E5%9B%BE%2CRt%E2%96%B3AB%E2%80%B2C%E2%80%B2%E6%98%AF%E7%94%B1Rt%E2%96%B3ABC%E7%BB%95%E7%82%B9A%E9%A1%BA%E6%97%B6%E9%92%88%E6%97%8B%E8%BD%AC%E5%BE%97%E5%88%B0%E7%9A%84%2C%E8%BF%9E%E6%8E%A5CC%E2%80%B2%E4%BA%A4%E6%96%9C%E8%BE%B9%E4%BA%8E%E7%82%B9E%2CCC%E2%80%B2%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%A4BB%E2%80%B2%E4%BA%8E%E7%82%B9F%EF%BC%8E%EF%BC%881%EF%BC%89%E8%8B%A5AC%3D3%2CAB%3D4%2C%E6%B1%82CC%E2%80%B2%EF%BC%9ABB%E2%80%B2)
如图,Rt△AB′C′是由Rt△ABC绕点A顺时针旋转得到的,连接CC′交斜边于点E,CC′的延长线交BB′于点F.(1)若AC=3,AB=4,求CC′:BB′
如图,Rt△AB′C′是由Rt△ABC绕点A顺时针旋转得到的,连接CC′交斜边于点E,CC′的延长线交BB′于点F.
(1)若AC=3,AB=4,求CC′:BB′ (2)证明:△ACE∽△FBE;
(3)设∠ABC=α,∠CAC′=β,试探索α、β满足什么关系时,△ACE与△FBE是全等三角形,并说明理由.
如图,Rt△AB′C′是由Rt△ABC绕点A顺时针旋转得到的,连接CC′交斜边于点E,CC′的延长线交BB′于点F.(1)若AC=3,AB=4,求CC′:BB′
(1)证明:∵Rt△AB′C′是由Rt△ABC绕点A顺时针旋转得到的
∴AC=AC′AB=AB′∠CA C′=∠B AB′
∴AC/AB=AC′/AB′
∴△AC C′∽△AB B′;
(1)证明:∵Rt△AB′C′是由Rt△ABC绕点A顺时针旋转得到的
∴AC=AC′AB=AB′∠CA C′=∠B AB′
∴ACAB=AC′AB′
∴△AC C′∽△AB B′
∴CC′:BB′ =AC:AB=3:4
(2)证明:∵Rt△AB′C′是由Rt△ABC绕点A顺时针旋转得到的,
∴AC=AC′,AB=AB′,∠CAB=∠C′AB′,
∴∠CAC′=∠BAB′,
∴∠ABB′=∠AB′B=∠ACC′=∠AC′C,
∴∠ACC′=∠ABB′,
又∵∠AEC=∠FEB,
∴△ACE∽△FBE.
(3)当β=2α时,AC=BF.
理由:∵AC=AC′
∴∠AC C′=∠A C′C=
(180°-∠C AC′)÷2=90°-1/2β=90°-α,
∵∠BCE=∠ACB-∠A C C′=90°-(90°-α)=α,
∴∠BCE=∠ABC,
∴BE=CE.
∵△AC C′∽△AB B′,
∵∠ACE=∠ABF.
在△AEC和△FEB中,
∠ACE=∠ABF
BE=CE
∠AEC=∠FEB
∴△AEC≌△FEB(ASA),
∴AC=BF.