AM,CM,分别平分∠BAD,∠BCD,求证:∠M=二分之一(∠B +∠D)
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![AM,CM,分别平分∠BAD,∠BCD,求证:∠M=二分之一(∠B +∠D)](/uploads/image/z/6738855-15-5.jpg?t=AM%2CCM%2C%E5%88%86%E5%88%AB%E5%B9%B3%E5%88%86%E2%88%A0BAD%2C%E2%88%A0BCD%2C%E6%B1%82%E8%AF%81%EF%BC%9A%E2%88%A0M%3D%E4%BA%8C%E5%88%86%E4%B9%8B%E4%B8%80%EF%BC%88%E2%88%A0B+%2B%E2%88%A0D%EF%BC%89)
AM,CM,分别平分∠BAD,∠BCD,求证:∠M=二分之一(∠B +∠D)
AM,CM,分别平分∠BAD,∠BCD,求证:∠M=二分之一(∠B +∠D)
AM,CM,分别平分∠BAD,∠BCD,求证:∠M=二分之一(∠B +∠D)
连AC,则
证明:连接AC
∵ ∠B+∠ACB+∠BAC=180°
∠D+∠DAC+DCA=180°
∠M+∠MAC+∠MCA=180°
又 ∵AM,CM,分别平分∠BAD,∠BCD
∴∠MAB=∠MAD;∠MCD=∠MCB
∴∠M=180°-∠MAC-∠MCA
=180°-(∠BAC-∠MAB)-(∠ACD-∠MC...
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证明:连接AC
∵ ∠B+∠ACB+∠BAC=180°
∠D+∠DAC+DCA=180°
∠M+∠MAC+∠MCA=180°
又 ∵AM,CM,分别平分∠BAD,∠BCD
∴∠MAB=∠MAD;∠MCD=∠MCB
∴∠M=180°-∠MAC-∠MCA
=180°-(∠BAC-∠MAB)-(∠ACD-∠MCD)
2*∠M =2*(180°-(∠BAC-∠MAD)-(∠ACD-∠MCB))
=360°-2(∠BAC-∠MAD)-2(∠ACD-∠MCB)
=360°-2(∠BAC-∠MAD)-2(∠ACD-∠MCB)
=360°-∠BAC -(∠BAC-2∠MAD)-∠ACD -(∠ACD-2∠MCB)
=360°-∠BAC -∠DAC-∠ACD -∠ACB
=360°-(∠BAC +∠ACB)-(∠ACD +∠DAC)
=360°-(180°-∠B)-(180°-∠D)
=∠B+∠D
∴∠M=二分之一(∠B +∠D)
证毕
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