Ta có : \(MB=MC=\dfrac{1}{2}BC\left(gt\right)\)
\(\Rightarrow AM=BM=MC\)
Hay : \(\Delta ABM\) là tam giác đều
\(\Rightarrow\widehat{ABM}=60^o\)
Hay : \(\widehat{ABC}=60^o\) (M thuộc BC)
Xét \(\Delta ABC\) có :
\(\widehat{ABC}=\widehat{CAB}=\widehat{ACB}=180^o\)
Hay : \(90^o+60^o+\widehat{BCA}=180^o\)
\(\Rightarrow\widehat{ACB}=30^o\)