, 2018

Denote by $\omega_a, \omega_b, \omega_c$, and $\omega_d$ the circles $AI_bI_d, BI_aI_c, CI_bI_d$, and $DI_aI_c$, 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. $I_bI_d \perp AC$ and $I_aI_c \perp BD$.

Proof. Let the incircles of triangles ABC and ACD be tangent to the line AC at T and T', respectively. (See Figure 1.) We have $AT = \frac{1}{2}(AB + AC - BC)$ in triangle ABC, $AT' = \frac{1}{2}(AD + AC - CD)$ 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 following Figure.)

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 $AI_d I_b$, 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 $AI_dI_b$ lies in the angle domain $I_bAI_d$ in such a way that $\angle I_bAT = \angle O_aAI_d$. The points $I_b$ and $I_d$ are the incenters of triangles $ABC$ and $ACD$, so the lines $AI_b$ and $AI_d$ bisect the angles $\angle BAC$ and $\angle CAD$, respectively. Then
$$\begin{aligned} \angle O_a AD &= \angle O_a AI_d + \angle I_d AD = \angle I_b AT + \angle I_d AD \\ &= \frac{1}{2} \angle BAC + \frac{1}{2} \angle CAD = \frac{1}{2} \angle BAD, \end{aligned}$$
so $O_a$ lies on the angle bisector of $\angle BAD$, that is, on the 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 $I_bI_d$ 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 $O_aI_b = O_aI_d = O_aA = r_a$ and $O_cI_b = O_cI_d = O_cC = r_c$, and from $AC \parallel O_aO_c$ we can

see that
$$\begin{aligned} \frac{O_a X}{O_c X} &= \frac{O_a U}{O_c U} = \frac{r_a}{r_c} = \frac{O_a I_b}{O_c I_b} \\ &= \frac{O_a I_d}{O_c I_d} = \frac{O_a A}{O_c C} = \frac{O_a I}{O_c I}. \end{aligned}$$
So the points $X, U, I_b, I_d$ lie on the Apollonius circle of the points $O_a, O_c$ with ratio $r_a : r_c$. In this Apollonius circle $XU$ is the diameter, and the lines $IU$ and $IX$ are respectively the internal and external bisectors of $\angle O_a IO_c = \angle AIC$, 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 internal and external bisectors of $\angle I_b II_d = \angle BID$, 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 $\angle AIC$ and the external bisector of $\angle BID$. Therefore, the lines $IX$ and $IY$ respectively are internal and external bisectors of $\angle BID$, so they are perpendicular. $\square$