如图,AB>AC,AD⊥BC,AE平分∠BAC,求证:∠DAE=二分之一(∠C-∠B)
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![如图,AB>AC,AD⊥BC,AE平分∠BAC,求证:∠DAE=二分之一(∠C-∠B)](/uploads/image/z/5190735-39-5.jpg?t=%E5%A6%82%E5%9B%BE%2CAB%EF%BC%9EAC%2CAD%E2%8A%A5BC%2CAE%E5%B9%B3%E5%88%86%E2%88%A0BAC%2C%E6%B1%82%E8%AF%81%EF%BC%9A%E2%88%A0DAE%3D%E4%BA%8C%E5%88%86%E4%B9%8B%E4%B8%80%EF%BC%88%E2%88%A0C-%E2%88%A0B%EF%BC%89)
如图,AB>AC,AD⊥BC,AE平分∠BAC,求证:∠DAE=二分之一(∠C-∠B)
如图,AB>AC,AD⊥BC,AE平分∠BAC,求证:∠DAE=二分之一(∠C-∠B)
如图,AB>AC,AD⊥BC,AE平分∠BAC,求证:∠DAE=二分之一(∠C-∠B)
∵AD⊥BC
∴∠DAC=90°-∠C
∵∠BAC=180°-∠B-∠C
AE平分∠BAC
∴∠EAC=1/2(180°-∠B-∠C)=90°-1/2∠B-1/2∠C
∵∠DAE=∠EAC-∠DAC
∴∠DAE=90°-1/2∠B-1/2∠C-(90°-∠C)
=1/2∠C-1/2∠B
=1/2(∠C-∠B)
∠DAC=90°-∠C
∠BAC=180°-∠B-∠C
∠DAE=1/2∠BAC-∠DAC
=90°-1/2∠B-1/2∠C-(90°-∠C)
=1/2(∠C-∠B)
∵∠BAC=180°-(∠B+∠C)∴∠CAE=1/2∠A=90°-1/2(∠B+∠C)∵∠DAE=∠CAE-∠CAD又∵∠CAD=90°-∠C∴∠DAE=90°-1/2(∠B+∠C)-(90°-∠C)=1/2(∠C-∠B)
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