Problem 2274

Let $\angle C A B=\alpha$, $\angle A B C=\beta$, $\angle B C A=\gamma$. We start by showing that $A$, $B$, $I_{1}$ and $I_{2}$ are concyclic. Since $A I_{1}$ and $B I_{2}$ bisect $\angle C A B$ and $\angle A B C$, their extensions beyond $I_{1}$ and $I_{2}$ meet at the incenter $I$ of the triangle. The points $E$ and $F$ are on the circle with diameter $B C$, so $\angle A E F=\angle A B C$ and $\angle A F E=\angle A C B$. Hence the triangles $A E F$ and $A B C$ are similar with ratio of similitude $\frac{A E}{A B}=\cos \alpha$. Because $I_{1}$ and $I$ are their incenters, we obtain $I_{1} A=I A \cos \alpha$ and $I I_{1}=I A-I_{1} A=2 I A \sin ^{2} \frac{\alpha}{2}$. By symmetry $I I_{2}=2 I B \sin ^{2} \frac{\beta}{2}$. The law of sines in the triangle $A B I$ gives $I A \sin \frac{\alpha}{2}=I B \sin \frac{\beta}{2}$. Hence
$$I I_{1} \cdot I A=2\left(I A \sin \frac{\alpha}{2}\right)^{2}=2\left(I B \sin \frac{\beta}{2}\right)^{2}=I I_{2} \cdot I B .$$
Therefore $A$, $B$, $I_{1}$ and $I_{2}$ are concyclic, as claimed.

In addition $I I_{1} \cdot I A=I I_{2} \cdot I B$ implies that $I$ has the same power with respect to the circles $\left(A C I_{1}\right)$, $\left(B C I_{2}\right)$ and $\left(A B I_{1} I_{2}\right)$. Then $C I$ is the radical axis of $\left(A C I_{1}\right)$ and $\left(B C I_{2}\right)$; in particular $C I$ is perpendicular to the line of centers $O_{1} O_{2}$.

Now it suffices to prove that $C I \perp I_{1} I_{2}$. Let $C I$ meet $I_{1} I_{2}$ at $Q$, then it is enough to check that $\angle I I_{1} Q+\angle I_{1} I Q=90^{\circ}$. Since $\angle I_{1} I Q$ is external for the triangle $A C I$, we have
$$\angle I I_{1} Q+\angle I_{1} I Q=\angle I I_{1} Q+(\angle A C I+\angle C A I)=\angle I I_{1} I_{2}+\angle A C I+\angle C A I .$$
It remains to note that $\angle I I_{1} I_{2}=\frac{\beta}{2}$ from the cyclic quadrilateral $A B I_{1} I_{2}$, and $\angle A C I=\frac{\gamma}{2}$, $\angle C A I=\frac{\alpha}{2}$. Therefore $\angle I I_{1} Q+\angle I_{1} I Q=\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}=90^{\circ}$, completing the proof.