Ta có: \(\left\{{}\begin{matrix}BC=2.AM\left(GT\right)\\BC=2.BM\left(GT\right)\end{matrix}\right.\)
=> AM = BM
=> ΔABM cân tại M
\(\Rightarrow\widehat{BAM}=\widehat{B}=\frac{180^0-\widehat{AMB}}{2}\) (1)
Ta có: \(\left\{{}\begin{matrix}BC=2.AM\left(GT\right)\\BC=2.CM\left(GT\right)\end{matrix}\right.\)
=> AM = CM
=> ΔACM cân tại M
\(\Rightarrow\widehat{MAC}=\widehat{C}=\frac{180^0-\widehat{AMC}}{2}\) (2)
Từ (1) và (2)
\(\Rightarrow\widehat{BAM}+\widehat{MAC}=\frac{180^0-\widehat{AMB}}{2}+\frac{180^0-\widehat{AMC}}{2}=\frac{\left(180^0-\widehat{AMB}\right)+\left(180^0-\widehat{AMC}\right)}{2}=\frac{180^0-\widehat{AMB}+180^0-\widehat{AMC}}{2}=\frac{360^0-\left(\widehat{AMB}+\widehat{AMC}\right)}{2}\)
Mà: \(\widehat{AMB}+\widehat{AMC}=180^0\) (kề bù)
\(\Rightarrow\widehat{BAM}+\widehat{MAC}=\frac{360^0-180^0}{2}=\frac{180^0}{2}=90^0\)
Hay: \(\widehat{BAC}=90^0\)
=> ΔABC vuông tại A
P/s: Mình làm chi tiết ra cho bạn dễ hiểu đó!
