2. In an acute triangle $\triangle ABC$, $H$ is the orthocenter, and $O$ is the circumcenter. The perpendicular bisector of $CH$ intersects $AC$ and $BC$ at points $X$ and $Y$, respectively. The lines $XO$ and $YO$ intersect $AB$ at points $P$ and $Q$, respectively. If $XP + YQ = AB + XY$, find $\angle OHC$.
2. As shown in Figure 1, let $N$ and $M$ be the midpoints of $CH$ and $AC$, respectively.

From $\triangle CYN \sim \triangle COM$
$\Rightarrow \angle OCY = \angle MCN, \frac{CY}{CN} = \frac{CO}{CM}$
$\Rightarrow \triangle CYO \sim \triangle CNM$
$\Rightarrow \angle CYO = \angle CNM$.
Let $\angle BYQ = \beta$, and $R$ be the circumradius of $\triangle ABC$.
Since $MN$ is the midline of $\triangle AHC$, we have
$\angle CHA = 180^{\circ} - \beta$.
Thus, $\angle ABC = \beta = \angle BYQ$.
Therefore, $YQ = BQ$.
Similarly, $XP = AP$.
From $XP + YQ = AB + XY$, we get
$XY = QP$.
Since $XY \parallel PQ$, quadrilateral $QPYX$ is a parallelogram.

Hence, the distance from point $O$ to $XY$ is equal to the distance from $O$ to $AB$, and $HN = R \cos \angle ACB$. Therefore, $OH \parallel XY$.
Since $HN \perp XY$, it follows that $\angle OHC = 90^{\circ}$.