Question

Answer 1

a: Ta có: \(\hat{NAF}=\hat{CBE}\)

mà \(\hat{CBE}+\hat{CBA}=180^0\) (hai góc kề bù)

và \(\hat{NAF}+\hat{BAN}=180^0\) (hai góc kề bù)

nên \(\hat{CBA}=\hat{BAN}\)

b: TA có: \(\hat{CBA}=\hat{BAN}\)

mà \(\hat{BAN}+\hat{BAM}=180^0\) (hai góc kề bù)

nên \(\hat{CBA}+\hat{BAM}=180^0\)