Prove: As shown in the figure, draw the diameter $KOE$ of the circumcircle of $\triangle ABC$, and connect $BK, BE$. In the right triangle $\triangle EBK$, $\angle KEB = \frac{1}{2} \angle A, KE = 2R$, then
$$\begin{aligned} BK = IK \\ = 2R \sin \frac{A}{2}. \end{aligned}$$

In the right triangle $\triangle AIF$, $IF = r$, then
$$AI = \frac{r}{\sin \frac{A}{2}}.$$

Thus, $AI \cdot IK$
$$\begin{array}{l} = \frac{r}{\sin \frac{A}{2}} \cdot (2R \sin \frac{A}{2}) \\ = 2rR. \end{array}$$

Also,
$$\begin{array}{l} \because \angle KBI = \angle KBC + \angle CBI \\ \quad = \frac{1}{2}(\angle A + \angle B) = \angle KIB, \\ \therefore BK = IK. \end{array}$$

Connect $O, I$ to get the diameter $K_1K_2$. According to the power of a point theorem,
$$2rR = AI \cdot IK = K_1I \cdot IK_2.$$

Since $K_1I = R - OI, IK_2 = R + OI$,
$$\therefore 2rR = (R - OI)(R + OI) = R^2 - OI^2,$$

i.e., $OI^2 = R^2 - 2rR$.
Note: The equation $OI^2 = R^2 - 2rR$ is known as Euler's relation.