如图△ABC∠C=90°∠ABC的角平分线与∠BAC的外角平分线相交于E点求证角E=45°
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如图△ABC∠C=90°∠ABC的角平分线与∠BAC的外角平分线相交于E点求证角E=45°
如图△ABC∠C=90°∠ABC的角平分线与∠BAC的外角平分线相交于E点求证角E=45°
如图△ABC∠C=90°∠ABC的角平分线与∠BAC的外角平分线相交于E点求证角E=45°
在EA延长线上取一点D,CA延长线上取一点F
∠ABE=∠ABC/2
∠BAF=180-∠BAC
∠DAF=(180-∠BAC)/2=90-∠BAC/2
因为∠CAE=∠DAF(对顶角)
所以∠CAE=90-∠BAC/2
∠E=180-∠ABE-∠BAC-∠CAE
=180-∠ABC/2-∠BAC-(90-∠BAC/2)
=90-∠ABC/2-∠BAC/2
=90-(∠ABC+∠BAC)/2
因为∠ABC+∠BAC=180-∠C
所以∠E=90-(180-∠C)/2
=∠C/2
∠ C=90,因此∠E=45
在EA延长线上取一点D,CA延长线上取一点F
∠ABE=∠ABC/2
∠BAF=180-∠BAC
∠DAF=(180-∠BAC)/2=90-∠BAC/2
因为∠CAE=∠DAF(对顶角)
所以∠CAE=90-∠BAC/2
∠E=180-∠ABE-∠BAC-∠CAE
=180-∠ABC/2-∠BAC-(90-∠BAC/2)
=90-∠A...
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在EA延长线上取一点D,CA延长线上取一点F
∠ABE=∠ABC/2
∠BAF=180-∠BAC
∠DAF=(180-∠BAC)/2=90-∠BAC/2
因为∠CAE=∠DAF(对顶角)
所以∠CAE=90-∠BAC/2
∠E=180-∠ABE-∠BAC-∠CAE
=180-∠ABC/2-∠BAC-(90-∠BAC/2)
=90-∠ABC/2-∠BAC/2
=90-(∠ABC+∠BAC)/2
因为∠ABC+∠BAC=180-∠C
所以∠E=90-(180-∠C)/2
=∠C/2
∠ C=90,因此∠E=45°
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在EA延长线上取一点D,CA延长线上取一点F
∠ABE=∠ABC/2
∠BAF=180-∠BAC
∠DAF=(180-∠BAC)/2=90-∠BAC/2
因为∠CAE=∠DAF(对顶角)
所以∠CAE=90-∠BAC/2
∠E=180-∠ABE-∠BAC-∠CAE
=180-∠ABC/2-∠BAC-(90-∠BAC/2)
=90-∠A...
全部展开
在EA延长线上取一点D,CA延长线上取一点F
∠ABE=∠ABC/2
∠BAF=180-∠BAC
∠DAF=(180-∠BAC)/2=90-∠BAC/2
因为∠CAE=∠DAF(对顶角)
所以∠CAE=90-∠BAC/2
∠E=180-∠ABE-∠BAC-∠CAE
=180-∠ABC/2-∠BAC-(90-∠BAC/2)
=90-∠ABC/2-∠BAC/2
=90-(∠ABC+∠BAC)/2
因为∠ABC+∠BAC=180-∠C
所以∠E=90-(180-∠C)/2
=∠C/2
∠ C=90,因此∠E=45
收起
在EA延长线上取一点D,CA延长线上取一点F
∠ABE=∠ABC/2
∠BAF=180-∠BAC
∠DAF=(180-∠BAC)/2=90-∠BAC/2
因为∠CAE=∠DAF(对顶角)
所以∠CAE=90-∠BAC/2
∠E=180-∠ABE-∠BAC-∠CAE
=180-∠ABC/2-∠BAC-(90-∠BAC/2)
=90-∠A...
全部展开
在EA延长线上取一点D,CA延长线上取一点F
∠ABE=∠ABC/2
∠BAF=180-∠BAC
∠DAF=(180-∠BAC)/2=90-∠BAC/2
因为∠CAE=∠DAF(对顶角)
所以∠CAE=90-∠BAC/2
∠E=180-∠ABE-∠BAC-∠CAE
=180-∠ABC/2-∠BAC-(90-∠BAC/2)
=90-∠ABC/2-∠BAC/2
=90-(∠ABC+∠BAC)/2
因为∠ABC+∠BAC=180-∠C
所以∠E=90-(180-∠C)/2
=∠C/2
∠ C=90,因此∠E=45°
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由三角形内角和=180°可知,
∠DAC=∠ABC+∠C;∠C=∠CAB+∠CBA=90°
由已知:∠EAC=1/2∠DAC=1/2∠ABC+1/2∠C;且∠ABE=1/2∠ABC,
∠E=180°-∠EAC-∠CAB-∠ABE
=180°-(1/2∠ABC+1/2∠C)-∠CAB-1/2∠ABC
=180°-∠ABC-∠CAB-1/2∠C
=180...
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由三角形内角和=180°可知,
∠DAC=∠ABC+∠C;∠C=∠CAB+∠CBA=90°
由已知:∠EAC=1/2∠DAC=1/2∠ABC+1/2∠C;且∠ABE=1/2∠ABC,
∠E=180°-∠EAC-∠CAB-∠ABE
=180°-(1/2∠ABC+1/2∠C)-∠CAB-1/2∠ABC
=180°-∠ABC-∠CAB-1/2∠C
=180°-90°-1/2∠C
=90°-(1/2)·90°【
?t=1332928704312】绘图地址
=45°
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