2. In the circle $\odot O$ with radius 1, there is an acute $\triangle A B C$ inscribed, $H$ is the orthocenter of $\triangle A B C$, and the angle bisector $A L$ is perpendicular to $O H$. Then $B C=$ $\qquad$ .
2. $\sqrt{3}$.

As shown in Figure 1, extend $AO$ to intersect $\odot O$ at point $N$, connect $AH$ and extend it to intersect $BC$ at point $H_{1}$, and connect $CN$.
In Rt $\triangle CAN$ and Rt $\triangle AH_{1}B$,
$\angle ANC=\angle ABC$.
Thus, $\angle CAN=\angle BAH_{1}$.
Since $AL$ is the angle bisector of $\triangle ABC$,
$\angle OAP=\angle HAP$.
Given $AP \perp OH$, we have $AH=AO=1$.
Extend $BO$ to intersect $\odot O$ at point $M$, and connect $AM$, $CM$,

and $CH$.
It is easy to see that quadrilateral $AMCH$ is a parallelogram.
Thus, $CM=AH=AO=1$.
Since $BM=2$, in Rt $\triangle MBC$, by the Pythagorean theorem,
$$
BC=\sqrt{2^{2}-1^{2}}=\sqrt{3} .
$$