If $AB$ and $CD$ are parallel, then obvious. So, let $AB$ and $CD$ intersect in $G$, first circle with center $E$ touch $CD$ in $I$, second with center $F$ touch $AB$ in $H$. Then $\sin{\angle BGC} = \frac{HF}{FC+CG}=\frac{EI}{EB+BG} \to \frac{HF}{CG}=\frac{EI}{BG}$ $\frac{AG}{GD}=\frac{2EI+BG}{2HF+CG}=\frac{BG}{CG}$ so $BC$ and $AD$ are parallel. Let $BC$ and $AD$ are parallel, and $H`$ on $AB$ such that $H`F$ perpendicular to $AB$ Easy to prove, that $ \frac{EI}{H`F}=\frac{BG}{CG}=\frac{AG}{GD}=\frac{2EI+BG}{2CF+CG} \to H`F=CF$

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Dear Mathlinkers, already posted https://artofproblemsolving.com/community/c6h60795p366610 Sincerely Jean-Louis

Dear Mathlinkers, see also http://jl.ayme.pagesperso-orange.fr/Docs/Geometrie%20enfin%20libre%201.pdf p. 9-10. Sincerely Jean-Louis

It was also Poland 2018 second stage P4