Serbian Mathematical Olympiad

Without loss of generality, let rays $AB$ and $DC$ intersect at point $P$, and rays $AD$ and $BC$ intersect at point $Q$. Denote the center of circle $\omega_a$ as $I_a$, and define points $I_b, I_c, I_d$ similarly. Let the quadrilateral formed by the four tangent lines be $A'B'C'D'$ (line $A'B'$ is the common external tangent to $\omega_a$ and $\omega_b$, and similarly for the other three sides).

For quadrilateral $ABCD$, denote $\angle A = \alpha$, $\angle D = \delta$. By the cyclic properties of quadrilaterals $ABCD$ and $AI_aI_dD$, we have angle relations: $\angle QCD = \alpha$, $\angle UI_aA = \angle ADI_d = \delta/2$, $\angle VI_dD = \angle DAI_a = \alpha/2$. Consequently, $\angle CQD = \delta - \alpha$ and $\angle PUV = \angle PVU = (\alpha+\delta)/2$. In particular, $\alpha < \delta$, so point $X_{ad}$ lies on ray $AD$ and $\angle DX_{ad}I_d = \angle ADV - \angle PVU = (\delta - \alpha)/2$. This equality implies that line $I_aI_d$ is parallel to the angle bisector of $\angle CQD$.
Since line $A'D'$ is symmetric to line $AD$ with respect to the line of centers $I_aI_d$, we obtain that $A'D' \parallel BC$, and line $A'D'$ passes through point $X_{ad}$. Defining points $X_{ab}, X_{bc}, X_{cd}$ similarly, we find these points lie on lines $A'B', B'C', C'D'$ which are parallel to the sides of quadrilateral $ABCD$.
As established, line $I_aI_d$ is parallel to the bisector of $\angle CQD$. Similarly, line $I_bI_c$ is parallel to this bisector, while lines $I_aI_b$

and $I_c I_d$ are parallel to the bisector of $\angle BPC$. Since $\angle PUV = \angle PVU$, the bisector of $\angle BPC$ is perpendicular to line $I_a I_d$. Thus, adjacent sides of quadrilateral $I_a I_b I_c I_d$ are perpendicular, making it a rectangle. Therefore, it is cyclic - denote its circumcircle by $\omega$, with center at the intersection of its diagonals.
It remains to prove that the centers of circles $\omega$, $\Omega$ and $\Gamma$ are collinear. We'll show these three circles share a radical axis containing points $X_{ab}, X_{bc}, X_{cd}, X_{da}$ ($\star$).
Let $\gamma$ be the circumcircle of quadrilateral $AI_a I_d D$. Then point $X_{ad}$ is the radical center of circles $\gamma$, $\omega$ and $\Omega$, lying on two of their radical axes. Hence, the radical axis of $\omega$ and $\Omega$ passes through $X_{ad}$, and similarly through $X_{ab}, X_{bc}, X_{cd}$. Thus these four points are collinear.
Let line $B'C'$ intersect side $AB$ at $S$ and side $CD$ at $T$. Since $B'C' \parallel AD$, we have $\angle BST = \angle BAD = 180^{\circ} - \angle BCT$, making quadrilateral $BCTS$ cyclic. As $C'D' \parallel AB$ and $A'B' \parallel CD$, by Thales' theorem:
$$\frac{X_{bc}B'}{X_{bc}T} = \frac{X_{bc}X_{ab}}{X_{bc}X_{cd}} = \frac{X_{bc}S}{X_{bc}C'}$$
From these ratios and the cyclicity of $BCTS$, we obtain $X_{bc}B' \cdot X_{bc}C' = X_{bc}S \cdot X_{bc}T = X_{bc}B \cdot X_{bc}C$, showing equal power of $X_{bc}$ with respect to $\Omega$ and $\Gamma$. Analogous reasoning applies to $X_{ab}, X_{ad}, X_{cd}$, proving claim ($\star$) as required.