1. **Identify the Midpoint $E$ of $AB$:**
Let $E$ be the midpoint of $AB$. Since $\ell$ is tangent to both circles $\Gamma_1$ and $\Gamma_2$ at points $A$ and $B$ respectively, and $E$ is the midpoint, we have:
$$|EA| = |EB|$$

2. Radical Axis and Perpendicular Lines:
The radical axis of two intersecting circles is the line that is perpendicular to the line joining their centers and passes through the points of intersection. Since $E$ is equidistant from $A$ and $B$, it lies on the radical axis of $\Gamma_1$ and $\Gamma_2$. This radical axis is the line $CD$.

3. Angles and Parallel Lines:
Since $AC$ and $BD$ are perpendicular to $\ell$, they are parallel to each other:
$$AC \parallel BD$$
This implies that the angles formed by these lines with the line $CD$ are equal:
$$\angle ACE = \angle BDE$$

4. Congruent Triangles:
Consider the triangles $\triangle ACE$ and $\triangle BDE$. Since $E$ is the midpoint of $AB$ and $AC \parallel BD$, we have:
$$\angle CEA = \angle DEB$$
Additionally, since $AC \parallel BD$, the corresponding angles are equal:
$$\angle ACE = \angle BDE$$
Therefore, the triangles $\triangle ACE$ and $\triangle BDE$ are congruent by the Angle-Angle (AA) criterion.

5. **Midpoint of $CD$:**
Since $\triangle ACE$ and $\triangle BDE$ are congruent, $E$ is also the midpoint of $CD$. This implies that $AB$ and $CD$ bisect each other at $E$.

6. Conclusion:
Since $AB \parallel CD$ and $AC \parallel BD$, and both pairs of opposite sides are equal in length, $ABCD$ is a parallelogram.

$\blacksquare$