Xét ΔABC có \(\hat{ABC}+\hat{ACB}+\hat{BAC}=180^0\)
=>\(\hat{ABC}+\hat{ACB}=180^0-\hat{BAC}\)
=>\(2\left(\hat{OBC}+\hat{OCB}\right)=180^0-\hat{BAC}\)
=>\(\hat{OBC}+\hat{OCB}=90^0-\frac12\cdot\hat{BAC}\)
Xét ΔBOC có \(\hat{OBC}+\hat{OCB}+\hat{BOC}=180^0\)
=>\(\hat{BOC}=180^0-90^0+\frac12\cdot\hat{BAC}=90^0+\frac12\cdot\hat{BAC}\) (1)
Ta có: Ay là phân giác của góc xAC
=>\(\hat{xAy}=\frac12\cdot\hat{xAC}=\frac12\left(180^0-\hat{BAC}\right)=90^0-\frac12\cdot\hat{BAC}\)
Ta có: \(\hat{xAy}+\hat{yAB}=180^0\) (hai góc kề bù)
=>\(\hat{yAB}=180^0-90^0+\frac12\cdot\hat{BAC}=90^0+\frac12\cdot\hat{BAC}\) (2)
Từ (1),(2) suy ra \(\hat{BOC}=\hat{yAB}\)
