Example 2 In $\triangle A B C$, $\angle A B C=60^{\circ}, O, H$ are the circumcenter and orthocenter of $\triangle A B C$ respectively. Points $D, E$ lie on sides $B C, A B$ respectively, such that $B D=B H, B E=B O$. Given that $B O=1$. Find the area of $\triangle B D E$. ${ }^{[1]}$
(2008, International Youth Math City Invitational Competition)
Solve As shown in Figure 2, draw the diameter $A M$ of the circumcircle of $\triangle A B C$, and connect $C M, B M, C H$.

Since $H$ is the orthocenter of $\triangle A B C$, we have
$$
\begin{array}{l}
B H \perp A C, \\
C H \perp A B .
\end{array}
$$

Since $A M$ is the diameter, we get $M C \perp A C, M B \perp A B$.
Therefore, $B H \parallel M C, C H \parallel M B$.
Thus, quadrilateral $B H C M$ is a parallelogram
$$
\Rightarrow B H=M C \text {. }
$$

Since $O M=O C, \angle A M C=\angle A B C=60^{\circ}$, we know that $\triangle O M C$ is an equilateral triangle. Hence, $M C=O C$.
Therefore, $B D=B H=M C=O C=B O=B E$.
Thus, $\triangle B D E$ is also an equilateral triangle.
Given $B O=1$, we have $S_{\triangle B D E}=\frac{\sqrt{3}}{4} B D^{2}=\frac{\sqrt{3}}{4}$.