Taiwan IMO Selection Camp

Let $\angle CAB = \alpha, \angle ABC = \beta, \angle BCA = \gamma$. We first prove that the four points $A, B, I_1, I_2$ are concyclic. Since $AI_1$ and $BI_2$ bisect $\angle CAB$ and $\angle ABC$ respectively, their extensions meet at the incenter $I$ of $\triangle ABC$. The two points $E, F$ lie on the circle with diameter $BC$, so we have $\angle AEF = \angle ABC$ and $\angle AFE = \angle ACB$. Hence $\triangle AEF$ is similar to $\triangle ABC$, with ratio of similarity $\frac{AE}{AB} = \cos \alpha$. Since $I_1$ and $I$ are the incenters of these two triangles respectively, we know that $I_1A = IA \cos \alpha$, and $II_1 = IA - I_1A = 2IA \sin^2 \frac{\alpha}{2}$. By symmetry we know that $II_2 = 2IB \sin^2 \frac{\beta}{2}$. According to

the Law of Sines, in $\triangle ABI$ we have $IA \sin \frac{\alpha}{2} = IB \sin \frac{\beta}{2}$, so we obtain
$$II_1 \cdot IA = 2\left(IA \sin \frac{\alpha}{2}\right)^2 = 2\left(IB \sin \frac{\beta}{2}\right)^2 = II_2 \cdot IB.$$
Thus it is proved that $A, B, I_1, I_2$ are concyclic.
From $II_1 \cdot IA = II_2 \cdot IB$ we can also obtain: for the circles $(ACI_1)$, $(BCI_2)$ and $(ABI_1I_2)$, the point $I$ has the same power of a point (here $(ACI_1)$ refers to the circle passing through these three points). Then $CI$ is the radical axis of the circles $(ACI_1)$ and $(BCI_2)$; from this we obtain that $CI$ is perpendicular to the line $O_1O_2$ joining the centers of these two circles.
Now it suffices to prove that $CI \perp I_1I_2$. Let $CI$ meet $I_1I_2$ at point $Q$, so we check whether $\angle II_1Q + \angle I_1IQ = 90^\circ$ holds. Since $\angle I_1IQ$ is an exterior angle of $\triangle ACI$, we obtain
$$\begin{aligned} \angle II_1Q + \angle I_1IQ &= \angle II_1Q + (\angle ACI + \angle CAI) \\ &= \angle II_1I_2 + \angle ACI + \angle CAI. \end{aligned}$$
Since $A, B, I_1, I_2$ are concyclic, we obtain $\angle II_1I_2 = \frac{\beta}{2}$, and also $\angle ACI = \frac{\gamma}{2}, \angle CAI = \frac{\alpha}{2}$, so $\angle II_1Q + \angle I_1IQ = \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = 90^\circ$, which completes the proof.