Let $E$ be the midpoint of $AB$. Since $|EA|=|EB|$, $E$ lies on the radical axis of $\Gamma_1, \Gamma_2$, which is the line $CD$. Then $\angle CEA = \angle DEB$ and moreover, since $AC \parallel BD$, $\angle ACE = \angle BDE$. Thus the triangles $ACE,BDE$ are congruent, i.e. $E$ is also the midpoint of $CD$. This proves the claim. (We see that the condition $AC \perp \ell \perp BD$ can be replaced by $AC \parallel BD$.)

Trivial. We only need $AC \parallel BD$ (which holds here as they are perpendicular to a common line). If $MN \cap AB = E$, then $\angle AME = \angle ANM$ by tangency and so $\triangle AME \sim \triangle ANM$, implying $EA^2 = EM \cdot EN$. Analogously $EB^2 = EM \cdot EN$, therefore $EA = EB$. Now with $AC \parallel BD$ we have $\angle ACE = \angle BDE$, so $\triangle ACE \cong \triangle BDE$, implying $AC = BD$, which together with $AC \parallel BD$ implies that $ABCD$ is a parallelogram.

It’s well-known $MN$ passes through the midpoint of $AB$ (e.g. power bash) so now as $AC\parallel BD$ and $CD$ passes through the midpoint of $AB$ we’re done.

Isn't this simple enough? Let $E$ be the midpoint of $AB$, from radical axis it is also $MN \cap AB$. Now triangles $AEC, BED$ are congruent, so $E$ is the midpoint of $CE$ as well; thus $ACBD$ is a parallelogram.

Attachments:

erm what the sigma? this is skibidi-trivial! this is because the radical axis bisects the common tangent!