, 2013

Let $p$ be the line tangent at $D$ to the circumcircle $\Gamma$ of $ABCD$. Since $C$ is the midpoint of the arc $BD$, we have $\angle(CD, p) = \angle CAD = \angle BAC = \angle BDC$, and we see that $p$ is tangent to $\omega$. Similarly, if $E$ is the midpoint of the arc $DA$ of $\Gamma$, then $p$ is tangent to the circle $\omega'$ centered at $E$ tangent to $DA$. Thus the line $q$ symmetric to $p$ with respect to $CE$ is tangent to $\omega$ and $\omega'$ (Fig. 2). However, the well-known relations $CD = CI$ and $ED = EI$ imply that $D$ and $I$ are symmetric with respect to $CE$. Hence $I$ lies on $q$ and it remains to show that $q \parallel AB$. This follows from
$$\angle(q, IC) = \angle(CD, p) = \angle CAD = \angle BAC$$
(all angles here are directed).

Fig. 2

Let $q$ denote the line through $I$ parallel to $AB$. $P$ is the orthogonal projection of $C$ on $q$, $M$ is the midpoint of $BD$ (Fig. 3). We have
$$\angle CIP = \angle CAB = \angle CDB = \angle CDM$$
Applying the well-known relation $CD = CI$, we conclude right triangles $CDM$ and $CIP$ are congruent, so $CM = CP$. The conclusion follows.

Fig. 3