18. $O$ is the circumcenter of $\triangle A D$, $X, Y$ are points on $A C, A B$, respectively, such that $B X$ and $C Y$ intersect at $O$. Also, $\angle B A C=\angle A Y X=\angle X Y C$. Try to find their common value.
18. It is easy to prove that $O$ is inside $\triangle ABC$. Connect $AO$ and extend it to intersect $BC$ at $Z$. By Ceva's Theorem,
$$
\frac{A Y \cdot B Z \cdot C X}{Y B \cdot Z C \cdot X A}=1.
$$

Since $YX$ bisects $\angle AYC$, we have $\frac{A Y}{Y C}=\frac{A X}{X C}$. Therefore, $\frac{B Y}{Y C}=\frac{B Z}{Z C}$, which means $YZ$ bisects $\angle BYC$. Let $\angle BAO=\alpha, \angle ACO=\beta, \angle CBO=\gamma$, then $\alpha+\beta+\gamma=90^{\circ}$. Since $\angle AYZ=\angle XYC=\alpha+\beta, \angle CYZ=\gamma$, thus $BYOZ$ is a cyclic quadrilateral, 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 means $3\alpha+\beta=90^{\circ}$. Solving the system of equations, we get $\beta=30^{\circ}, \alpha=20^{\circ}$, thus $\angle BAC=50^{\circ}$.