2. In $\triangle A B C$, $\angle A B C=60^{\circ}$, points $O$ and $H$ are the circumcenter and orthocenter of $\triangle A B C$, respectively. Points $D$ and $E$ are on sides $B C$ and $A B$, respectively, such that $B D=B H, B E=B O$. Given that $B O=1$, find the area of $\triangle B D E$.
2. As shown in Figure 12, draw the diameter $A F$ of the circumcircle of $\triangle A B C$, and connect $C F$, $B F$, and $C H$.

Since $B H \perp A C$ and $F C \perp A C$, we have $B H \parallel F C$.
Similarly, $C H \parallel F B$.
Therefore, quadrilateral $B H C F$ is a parallelogram.
Furthermore, since $F O = C O$ and $\angle A F C = \angle A B C = 60^{\circ}$, $\triangle F O C$ is an equilateral triangle.
Thus, $B D = B H = C F = C O = B O = B E$.
Hence, $\triangle B D E$ is also an equilateral triangle.
Given that $B O = 1$, we have $S_{\triangle B D E} = \frac{\sqrt{3}}{4}$.