Given that $O$ is the circumcenter of $\triangle A B C$, $X$ and $Y$ are points on $A C$ and $A B$ respectively, such that $B X$ and $C Y$ intersect at $O$, and $\angle B A C = \angle A Y X = \angle X Y C$. Find the measure of $\angle B A C$.
It is known that $O$ is inside $\triangle ABC$. Connect $AO$ and extend it to intersect $BC$ at $Z$. By Ceva's Theorem, we have
$$
\frac{AY}{YB} \cdot \frac{BZ}{ZC} \cdot \frac{CX}{XA}=1
$$

Since $YX$ bisects $\angle AYC \Rightarrow \frac{AY}{YC}=\frac{AX}{XC}$
Therefore, $YZ$ bisects $\angle BYC$
Let $\angle BAO=\alpha, \angle ACO=\beta, \angle CBO=\gamma$, then
$$
\alpha+\beta+\gamma=90^{\circ}
$$

Since $\angle AYX=\angle XYC=\alpha+\beta$
Therefore, $\angle CYZ=\gamma$, which means $\angle CBO=\angle CYZ$
Thus, $B, Y, O, Z$ are concyclic
Therefore, $\angle YZO=\alpha$
Also, the sum of the interior angles of $\triangle AYC$ is $3\alpha+4\beta=180^{\circ}$
The sum of the interior angles of $\triangle AYZ$ is $4\alpha+2\beta+\gamma=180^{\circ}$, which means $3\alpha+\beta=90^{\circ}$

Therefore, $\alpha=20^{\circ}, \beta=30^{\circ}$
Therefore, $\angle BAC=50^{\circ}$