China Southeastern Mathematical Olympiad

As shown in Fig. 2, let $OI=d$, $R$, $r$ be the circumradius and incircle radius of $△ABC$. 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 $$PE=2R\sin \frac{\angle EDF}{2},DI=\frac{r}{\sin \frac{\angle EDF}{2}},$$ $$ID\cdot IP=IM\cdot IN=(R+d)(R-d)=R^2-d^2,$$ 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 $△DEF$ [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 $I$.