一、(20 points) As shown in Figure 5, in $\triangle ABC$, $\angle ABC=60^{\circ}$, $O$ and $H$ are the circumcenter and orthocenter of $\triangle ABC$ respectively. Points $D$ and $E$ are on sides $BC$ and $AB$ respectively, and $BD=BH$, $BE=BO$, $BO=a$. Find the area of $\triangle BDE$.
Construct the diameter $A F$ of the circumcircle of $\triangle A B C$, and connect $C O, C F, B F, C H$.
Since $B H \perp A C, F C \perp A C$, we have $B H / / F C$.
Similarly, $B F / / H C$.
Therefore, quadrilateral $B H C F$ is a parallelogram.
Given $F O=C O, \angle A F C=\angle A B C=60^{\circ}$
$\Rightarrow \triangle F O C$ is an equilateral triangle.
Thus, $B D=B H=C F=C O=B O=B E=a$
$\Rightarrow \triangle B D E$ is also an equilateral triangle.
Hence, $S_{\triangle B D E}=\frac{\sqrt{3}}{4} B D^{2}=\frac{\sqrt{3}}{4} a^{2}$.