China Southeastern Mathematical Olympiad

As shown in Fig. 2, let $OI = d$, $R$, $r$ be the circumradius and incircle radius of $\triangle ABC$. The point $K$ is the intersection of $AI$ and the circle $O$; then
$$KI = KB = 2R \sin \frac{\angle BAC}{2},$$
$$AI = \frac{r}{\sin \frac{\angle BAC}{2}}.$$

Fig. 2
Let the points $M, N$ be the intersections of the line $OI$ and the circle $O$; then
$$(R+d)(R-d) = IM \times IN = AI \times KI = 2Rr,$$
i.e. $R^2 - d^2 = 2Rr$.

Now draw the tangents $DE, DF$ from $D$ to the circle $I$. The points $E, F$ are on the circle $O$. Then $DI$ is the bisector of $\angle EDF$. It is enough to prove that $EF$ is tangent to the circle $I$.

Let $P$ be the intersection of the line $DI$ and the circle $O$. Then $P$ is the midpoint of the arc $EF$, and
$$PE = 2R \sin \frac{\angle EDF}{2}, \quad DI = \frac{r}{\sin \frac{\angle EDF}{2}},$$
$$ID \cdot IP = IM \cdot IN = (R+d)(R-d) = R^2 - d^2,$$
and so
$$PI = \frac{R^2 - d^2}{DI} = \frac{R^2 - d^2}{r} \cdot \sin \frac{\angle EDF}{2} = 2R \sin \frac{\angle EDF}{2} = PE.$$
As $I$ is on the bisector of $\angle EDF$, we can see that $I$ is the incenter of $\triangle DEF$ [because $\angle PEI = \angle PIE = \frac{1}{2}(180^\circ - \angle EPD) = \frac{1}{2}(180^\circ - \angle DFE) = \frac{\angle EDF + \angle DEF}{2}$, and $\angle PEF = \frac{\angle EDF}{2}$; so $\angle FEI = \frac{\angle DEF}{2}$]. $EF$ is tangent to the circle $I$.