Do I miss something or the problem is not true. Take any triangle $ABC$ with $BA\neq BC$, let $D$ be reflection of $B$ over $AC$. For this quadrilateral, the statement does not hold even though $ABCD$ has incircle and there is line $l$ parallel to $BD$ tangent to incircles of triangles $ABC,CDA$ by symmetry.

rafaello wrote: Do I miss something or the problem is not true. Take any triangle $ABC$ with $BA\neq BC$, let $D$ be reflection of $B$ over $AC$. For this quadrilateral, the statement does not hold even though $ABCD$ has incircle and there is line $l$ parallel to $BD$ tangent to incircles of triangles $ABC,CDA$ by symmetry. It's not circumscribed

Do you mean circumscribed quadrilateral as having circumcircle or what?

rafaello wrote: Do you mean circumscribed quadrilateral as having circumcircle or what? Yes, edited, sorry

We will consider the external tangent $\ell$ which is closer to $C$ and will prove that if $\ell \parallel BD$ then $\ell$ passes through the incenter of $BCD$. We fix the triangle $BCD$ and move the point $A$ along the arc $BD$. If $A$ coincide with the midpoint of the arc $BD$, then we can reflect $BC$ with the respect to the bisector of $BAC$ and $AD$ with respect to the bisector of $DAC$ and get the same line, which due to simple angle chasing will be parallel to $BD$ and tangent to both incircles. The images of points $B$ and $D$ after such reflections will coincide with the incenter of $BCD$ because of trillium theorem. Now it is easy to check, that if we move the point $A$ closer to $B$ then the left incircle will become smaller and the right one will become bigger and the common external tangent will be no longer parallel to $BD$. This completes the proof.

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