(1)如图所示,∠BAC=90°,AD⊥BC,垂足为D,BE平分∠ABC,交AC于E,交AD于F,试判断△AEF的形状,
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![](http://img.wesiedu.com/upload/e/b3/eb30909211d560b8acb4f7b87d8fcd7f.jpg)
(1)如图所示,∠BAC=90°,AD⊥BC,垂足为D,BE平分∠ABC,交AC于E,交AD于F,试判断△AEF的形状,并说明理由;
(2)如图所示,已知∠BAC=90°,AD⊥BC,垂足为D,AE=AF,试说明BE平分∠ABC.
![(1)如图所示,∠BAC=90°,AD⊥BC,垂足为D,BE平分∠ABC,交AC于E,交AD于F,试判断△AEF的形状,](/uploads/image/z/19594734-6-4.jpg?t=%EF%BC%881%EF%BC%89%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%EF%BC%8C%E2%88%A0BAC%3D90%C2%B0%EF%BC%8CAD%E2%8A%A5BC%EF%BC%8C%E5%9E%82%E8%B6%B3%E4%B8%BAD%EF%BC%8CBE%E5%B9%B3%E5%88%86%E2%88%A0ABC%EF%BC%8C%E4%BA%A4AC%E4%BA%8EE%EF%BC%8C%E4%BA%A4AD%E4%BA%8EF%EF%BC%8C%E8%AF%95%E5%88%A4%E6%96%AD%E2%96%B3AEF%E7%9A%84%E5%BD%A2%E7%8A%B6%EF%BC%8C)
(1)△AEF是等腰三角形,
理由如下:
∵BF平分∠ABC,![](http://img.wesiedu.com/upload/e/b3/eb30909211d560b8acb4f7b87d8fcd7f.jpg)
∴∠ABF=∠DBF,
又∵∠BAC=90°,AD⊥BC,
∴∠AFE=90°-∠ABF,∠DEB=90°-∠DBF,
∴∠AFE=∠DEB,
又∵∠DEB=∠AEF,
∴∠AEF=∠AFE,
∴△AEF是等腰三角形;
(2)证明:
∵∠BAC=90°,AD⊥BC,
∴∠AFE+∠ABF=90°,∠DEB+∠BED=90°,
∵AE=AF,
∴∠AFE=∠AEF,
∴∠ABF=∠DBF,
∴BF平分∠ABC.
理由如下:
∵BF平分∠ABC,
![](http://img.wesiedu.com/upload/e/b3/eb30909211d560b8acb4f7b87d8fcd7f.jpg)
∴∠ABF=∠DBF,
又∵∠BAC=90°,AD⊥BC,
∴∠AFE=90°-∠ABF,∠DEB=90°-∠DBF,
∴∠AFE=∠DEB,
又∵∠DEB=∠AEF,
∴∠AEF=∠AFE,
∴△AEF是等腰三角形;
(2)证明:
∵∠BAC=90°,AD⊥BC,
∴∠AFE+∠ABF=90°,∠DEB+∠BED=90°,
∵AE=AF,
∴∠AFE=∠AEF,
∴∠ABF=∠DBF,
∴BF平分∠ABC.
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