Đáp án:
Giải thích các bước giải:
Do `ABCD` là tứ giác
`=>\hat{A}+\hat{B}+\hat{ADC}+\hat{BCD}=360^0`
`=>\hat{A}+\hat{B}+2\hat{ODC}+2\hat{OCD}=360^0`
`=>2(\hat{ODC}+\hat{OCD})=360^0-(\hat{A}+\hat{B})`
`=>2(180^0-\hat{COD})=360^0-(\hat{A}+\hat{B})`
`=>360^0-2\hat{COD}=360^0-(\hat{A}+\hat{B})`
`=>2\hat{COD}=\hat{A}+\hat{B}`
`=>\hat{COD}=(\hat{A}+\hat{B})/2`
`=>đ.p.c.m`
