Cesko-Slovacko-Poljsko 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 ${\omega }^{\prime }$ centered at $E$ tangent to $DA$. Thus the line $q$ symmetric to $p$ with respect to $CE$ is tangent to $\omega$ and ${\omega }^{\prime }$ (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

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Alternative solution.

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