One, (50 points) As shown in the figure, $O$ is the circumcenter of $\triangle ABC$, and $X, Y$ are points on $AC, AB$ respectively, such that $BX$ and $CY$ intersect at point $O$, and $\angle BAC = \angle AYX = \angle XYC$. Find the size of $\angle BAC$.

保留源文本的换行和格式，直接输出翻译结果如下：

```
One, (50 points) As shown in the figure, $O$ is the circumcenter of $\triangle ABC$, and $X, Y$ are points on $AC, AB$ respectively, such that $BX$ and $CY$ intersect at point $O$, and $\angle BAC = \angle AYX = \angle XYC$. Find the size of $\angle BAC$.
```
I. Solution: Connect $AO$ and extend it to intersect $BC$ at $Z$, connect $YZ$. By Ceva's Theorem, we get $\frac{AY}{YB} \cdot \frac{BZ}{ZC} \cdot \frac{CX}{XA}=1$.
Since $XY$ bisects $\angle AYC$, we have $\frac{AY}{YC}=\frac{AX}{XC}$,
thus, $\frac{BY}{YC}=\frac{BZ}{CZ}$, hence $ZY$ bisects $\angle BYC$.
As shown in the figure,
let $\angle BAO=\alpha, \angle ACO=\beta, \angle CBO=\gamma$,
it is easy to see that $\alpha+\beta+\gamma=\frac{1}{2} \cdot 180^{\circ}=90^{\circ}$.
Since $\angle AYX=\angle XYC=\angle BAC=\alpha+\beta$, we get $\angle CYZ=\gamma$, thus $B, Y, O, Z$ are concyclic, so $\angle YZO=\alpha$, the sum of the interior angles of $\triangle AYC$ is $3\alpha+4\beta=180^{\circ}$, and the sum of the interior angles of $\triangle AYZ$ is $4\alpha+2\beta+\gamma=180^{\circ}$, which gives $3\alpha+\beta=90^{\circ}$, thus solving for $\alpha=20^{\circ}, \beta=30^{\circ}$, hence $\angle BAC=50^{\circ}$.