International Mathematical Olympiad

Denote by $\omega_{a}$, $\omega_{b}$, $\omega_{c}$ and $\omega_{d}$ the circles $A I_{b} I_{d}$, $B I_{a} I_{c}$, $C I_{b} I_{d}$, and $D I_{a} I_{c}$, let their centers be $O_{a}$, $O_{b}$, $O_{c}$ and $O_{d}$, and let their radii be $r_{a}$, $r_{b}$, $r_{c}$ and $r_{d}$, respectively.

Claim 1. $I_{b} I_{d} \perp AC$ and $I_{a} I_{c} \perp BD$.

Proof. Let the incircles of triangles $ABC$ and $ACD$ be tangent to the line $AC$ at $T$ and $T'$, respectively. (See the figure to the left.) We have $AT = \frac{AB + AC - BC}{2}$ in triangle $ABC$, $AT' = \frac{AD + AC - CD}{2}$ in triangle $ACD$, and $AB - BC = AD - CD$ in quadrilateral $ABCD$, so
$$AT = \frac{AC + AB - BC}{2} = \frac{AC + AD - CD}{2} = AT'.$$
This shows $T = T'$. As an immediate consequence, $I_{b} I_{d} \perp AC$.

The second statement can be shown analogously. $\square$

Claim 2. The points $O_{a}$, $O_{b}$, $O_{c}$ and $O_{d}$ lie on the lines $AI$, $BI$, $CI$ and $DI$, respectively.

Proof. By symmetry it suffices to prove the claim for $O_{a}$. (See the figure to the right above.)

Notice first that the incircles of triangles $ABC$ and $ACD$ can be obtained from the incircle of the quadrilateral $ABCD$ with homothety centers $B$ and $D$, respectively, and homothety factors less than $1$, therefore the points $I_{b}$ and $I_{d}$ lie on the line segments $BI$ and $DI$, respectively.

As is well-known, in every triangle the altitude and the diameter of the circumcircle starting from the same vertex are symmetric about the angle bisector. By Claim 1, in triangle $A I_{d} I_{b}$, the segment $AT$ is the altitude starting from $A$. Since the foot $T$ lies inside the segment $I_{b} I_{d}$, the circumcenter $O_{a}$ of triangle $A I_{d} I_{b}$ lies in the angle domain $I_{b} A I_{d}$ in such a way that $\angle I_{b} A T = \angle O_{a} A I_{d}$. The points $I_{b}$ and $I_{d}$ are the incenters of triangles $ABC$ and $ACD$, so the lines $A I_{b}$ and $A I_{d}$ bisect the angles $\angle BAC$ and $\angle CAD$, respectively. Then
$$\angle O_{a} A D = \angle O_{a} A I_{d} + \angle I_{d} A D = \angle I_{b} A T + \angle I_{d} A D = \frac{1}{2} \angle BAC + \frac{1}{2} \angle CAD = \frac{1}{2} \angle BAD,$$
so $O_{a}$ lies on the angle bisector of $\angle BAD$, that is, on line $AI$. $\square$

The point $X$ is the external similitude center of $\omega_{a}$ and $\omega_{c}$; let $U$ be their internal similitude center. The points $O_{a}$ and $O_{c}$ lie on the perpendicular bisector of the common chord $I_{b} I_{d}$ of $\omega_{a}$ and $\omega_{c}$, and the two similitude centers $X$ and $U$ lie on the same line; by Claim 2, that line is parallel to $AC$.

From the similarity of the circles $\omega_{a}$ and $\omega_{c}$, from $O_{a} I_{b} = O_{a} I_{d} = O_{a} A = r_{a}$ and $O_{c} I_{b} = O_{c} I_{d} = O_{c} C = r_{c}$, and from $AC \parallel O_{a} O_{c}$ we can see that
$$\frac{O_{a} X}{O_{c} X} = \frac{O_{a} U}{O_{c} U} = \frac{r_{a}}{r_{c}} = \frac{O_{a} I_{b}}{O_{c} I_{b}} = \frac{O_{a} I_{d}}{O_{c} I_{d}} = \frac{O_{a} A}{O_{c} C} = \frac{O_{a} I}{O_{c} I}.$$
So the points $X, U, I_{b}, I_{d}, I$ lie on the Apollonius circle of the points $O_{a}, O_{c}$ with ratio $r_{a} : r_{c}$. In this Apollonius circle $XU$ is a diameter, and the lines $IU$ and $IX$ are respectively the internal and external bisectors of $\angle O_{a} I O_{c} = \angle A I C$, according to the angle bisector theorem. Moreover, in the Apollonius circle the diameter $UX$ is the perpendicular bisector of $I_{b} I_{d}$, so the lines $IX$ and $IU$ are the internal and external bisectors of $\angle I_{b} I I_{d} = \angle B I D$, respectively.

Repeating the same argument for the points $B, D$ instead of $A, C$, we get that the line $IY$ is the internal bisector of $\angle A I C$ and the external bisector of $\angle B I D$. Therefore, the lines $IX$ and $IY$ respectively are the internal and external bisectors of $\angle B I D$, so they are perpendicular.