a) Ta có:
\(\widehat{MAN}=\widehat{MAB}+\widehat{BAC}+\widehat{NAC}\)
=> \(\widehat{MAN}=60+60+60\)
=>\(\widehat{MAN}=180\)
=> MAN là góc bẹt
=> M, A, N thẳng hàng
b) Ta có:
+) \(\widehat{MAC}=\widehat{MAB}+\widehat{BAC}\)
=> \(\widehat{MAC}=60+60=120\) (1)
+) \(\widehat{NAB}=\widehat{NAC}+\widehat{BAC}\)
=> \(\widehat{NAB}=60+60=120\) (2)
Từ (1) và (2) => \(\widehat{MAC}=\widehat{NAB}\)
Xét tam giác MAC và BAN có:
MA=AB
\(\widehat{MAC}=\widehat{NAB}\) (cmt)
NA=AC
=> Tam giác MAB = tam giác BAN ( c-g-c)
=> BN=CM (2 cạnh tương ứng)
c) Ta có: Tam giác MAB = tam giác BAN (câu b)
=> \(\widehat{MCA}=\widehat{NBA}\)( 2 góc tương ứng)
Mặt khác ta có:
\(\widehat{NOC}=180-\left[\left(\widehat{MCA}+\widehat{ACN}\right)+\left(\widehat{ANC}-\widehat{BAN}\right)\right]\)
=> \(\widehat{NOC}=180-\left[\left(\widehat{MCA}+60\right)+\left(60-\widehat{BNA}\right)\right]\)
=> \(\widehat{NOC}=180-\widehat{MCA}+60+60-\widehat{BAN}\)
\(\Rightarrow\widehat{NOC}=180-\left(60+60\right)\)
=>\(\widehat{NOC}=120\)
Ta lại có:
\(\widehat{NOC}+\widehat{BOC}=180\) (kề bù)
=> \(120+\widehat{BOC}=180\)
=> \(\widehat{BOC}=60\)
