6. In $\triangle A B C$, $A B=A C$, semicircle $O$ is tangent to $A B$ and $A C$, and the center $O$ is on side $B C$. A line is tangent to the semicircle and intersects $A B$ and $A C$ at points $E$ and $F$, respectively. Then $A O$ passes through the ( ) of $\triangle O E F$.
(A) circumcenter
(B) incenter
(C) centroid
(D) orthocenter
$$
\begin{array}{l}
\text { 6.A. } \\
\text { Let } \angle B=\angle C=\alpha, \\
\angle B E F=2 \beta, \angle C F E=
\end{array}
$$
$2 \gamma$. Since the sum of the interior angles of quadrilateral $B C F E$ is $360^{\circ}$, that is,
$$
\alpha+\alpha+2 \beta+2 \gamma=360^{\circ},
$$

hence $\alpha+\beta+\gamma=180^{\circ}$.
In $\triangle B O E$, it is easy to see that $\angle B O E=\gamma=\angle E F O$.
By the converse of the tangent-chord theorem, $O B$ is tangent to the circumcircle of $\triangle O E F$, and $O$ is the point of tangency. Since $A O \perp O B$, $A O$ passes through the center of the circumcircle of $\triangle O E F$.