Berkeley Math Circle

Solution:

It is enough to show that $DI^{2} - DO^{2} = EI^{2} - EO^{2}$. But if we let $R$ denote the circumradius, then by power of a point we have $R^{2} - DO^{2} = AD \cdot DB$, and $R^{2} - EO^{2} = AE \cdot EC$. Thus it suffices to prove
$$DI^{2} + AD \cdot DB = EI^{2} + AE \cdot EC \Longleftrightarrow DI^{2} - EI^{2} = AE \cdot EC - AD \cdot DB.$$
In the usual notation $a = BC$, $b = CA$, $c = AB$, the right-hand side is
$$AE \cdot EC - AD \cdot DB = (b - a)a - (c - a)a = a(b - c).$$
Now let the foot from $I$ to $BC$ be $K$; it's well-known that $BK = \frac{1}{2}(a - b + c)$ and $CK = \frac{1}{2}(a + b - c)$. So
$$\begin{aligned} DI^{2} - EI^{2} & = CI^{2} - BI^{2} \\ & = \left(IK^{2} + CK^{2}\right) - \left(IK^{2} + BK^{2}\right) \\ & = (CK - BK)(CK + BK) \\ & = a(c - b). \end{aligned}$$
This gives the desired equality, so we're done.