9.6. Given that the incircle of $\triangle ABC$ touches sides $AB, AC$ at points $X, Y$, and $K$ is the midpoint of the arc $\overparen{AB}$ (not containing point $C$) of the circumcircle of $\triangle ABC$, the line $XY$ bisects the segment $AK$. Find the size of $\angle BAC$.
9.6. Let $I$ be the incenter of $\triangle ABC$, and $XY$ intersects $AI$ at point $S$, $T$ is the midpoint of $AI$.
From the collinearity of $C$, $I$, and $K$, we get
$$
\begin{array}{l}
\angle AIK = \angle CAI + \angle ACI \\
= \angle BAI + \angle KAB = \angle KAI.
\end{array}
$$

Thus, $KT \perp AI$.
Since $XY \perp AI$, then $XY \parallel KT$. Because $XY$ passes through the midpoint of $AK$, $S$ is the midpoint of $AT$, so $\frac{AS}{SI} = \frac{1}{3}$.
Since $XS$ is the altitude from the right angle of $\triangle AI X$ to the hypotenuse $AI$, we have $XS = \sqrt{AS \cdot SI} = \sqrt{3} AS$.
Therefore, $\angle XAI = 60^\circ$, $\angle BAC = 120^\circ$.