As shown in Figure 4, connect $IF$, $IG$, $HF$, $HG$, and $FG$.
From the given, $DC \Perp_{2} HG$, $EB \Perp_{2} FH$.
Also, since $I$, $H$, $G$, and $F$ are concyclic,
$$\angle IGF = \angle IHF = \angle AED.$$
Similarly, $\angle IFG = \angle ADE$.
Thus, $\triangle IFG \backsim \triangle ADE$
$$\Rightarrow IF : FG : GI = AD : DE : EA.$$

For the cyclic quadrilateral $IFGH$, by Ptolemy's theorem, we have
$$\begin{array}{l} IF \cdot HG + FG \cdot IH = GI \cdot FH \\ \Rightarrow AD \cdot 2HG + DE \cdot 2IH = EA \cdot 2FH \\ \Rightarrow AD \cdot DC + DE \cdot 2IH = EA \cdot EB \\ \Rightarrow \left(R^2 - OE^2\right) + \left(EI^2 - DI^2\right) \\ = R^2 - OD^2 \quad (R \text{ is the radius of } \odot O) \\ \Rightarrow OD^2 - OE^2 = DI^2 - EI^2 \\ \Rightarrow OI \perp DE. \end{array}$$