[Toán 7] CM tam giác đều

A

ampm

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H

hiensau99

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a, + $\triangle ABC $ có $\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^o$. Hay $60^o+\widehat{BAC}=180^o \Longrightarrow \widehat{BAC}=120^o$

+ AD là tia phân giác $ \widehat{BAC} \Longrightarrow \widehat{A_2}=\widehat{A_3}=120^o:2= 60^o $

+ Ta có $\widehat{BAC} +\widehat{A_1} = 180^o $ (kề bù). Hay $120^o+\widehat{A_1}=180^o \Longrightarrow \widehat{A_1}=60^o $

+ Ta có $\widehat{A_1}=60^o=\widehat{A_4}$ (đồi đỉnh)
$ \Longrightarrow \widehat{A_1}=\widehat{A_4}= \widehat{A_2}=\widehat{A_3}= 60^o$

+ CM: $\triangle NAC= \triangle DAC$ (gcg)
$\Longrightarrow NA=AD$ (2 cạnh tương ứng) (1)

+ CM: $\triangle AMB= \triangle ADB$ (gcg)
$\Longrightarrow AD=AM$ (2 cạnh tương ứng) (2)

+ Từ (1) và (2) $\Longrightarrow AM=AN $ (đpcm)

b, + ta có $\widehat{MAN}=\widehat{BAC}=120^o$ (đối đỉnh)

+ ta có $\widehat{MAO}=\widehat{A_4}+\widehat{A_3}= 60^o.2=120^o$

+ Ta có $\widehat{NAO}=\widehat{A_1}+\widehat{A_2}= 2.60^o= 120^o$

+ Theo phần a: $AN=AO \Longrightarrow \triangle NAO$ cân ở A $\Longrightarrow \widehat{ANO}=\widehat{AON}= (180^o-\widehat{NAO}):2= 30^o $

+ Theo phần a: $AN=AM \Longrightarrow \triangle NAM$ cân ở A $\Longrightarrow \widehat{AMN}=\widehat{ANM}= (180^o-\widehat{NAM}):2= 30^o $

+ Theo phần a: $AM=AO \Longrightarrow \triangle MAO$ cân ở A $\Longrightarrow \widehat{AMO}=\widehat{AOM}= (180^o-\widehat{MAO}):2= 30^o $

+ Ta có $\widehat{MNO}=\widehat{MNA}+\widehat{ANO}=30^o.2=60^o$

+ Ta có $\widehat{NMO}=\widehat{NMA}+\widehat{AMO}=30^o.2=60^o$

+ $\triangle MNO$ có $\widehat{MNO}=\widehat{NMO}=60^o$
$\Longrightarrow \triangle MNO $ đều (đpcm)
 
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