a/
Đặt `\hat{ABC}=\hat{A}`
Ta có: `\hat{B}=hat{A_1,C}=\hat{A_2}=>\hat{B}+\hat{C}=\hat{A_1}+\hat{A_2}=\hat{A}`
Do `\hat{B}+\hat{C}+\hat{A}=180^0` nên `\hat{A}=90^0`
b/
Trên tia `MA` lấy điểm D sao cho `MD=\frac{BC}{2}` thì D nằm giữa `M` và `A`
Ta có: `\hat{BAM}<\hat{BDM},\hat{CAM}<\hat{CDM}=>\hat{BAM}+\hat{CAM}
<\hat{BDM}+\hat{CDM}`
`=>\hat{BAC}<\hat{BDC}=90^0`
c/
Ta có:
`AM<BM`
`=>\hat{B}<\hat{A_1};\hat{C}<\hat{A_2}`
`=>\hat{B}+\hat{C}<\hat{A}`, mà `\hat{A}+\hat{B}+\hat{B}=180^0`
`=>\hat{A}>90^0`