Let $O$ be the center of the triangle. Let $M$ and $N$ be intersection of $\ell$ with $AB$ and $BC$ respectively, finally let $M'$ and $N'$ be points symmetric to $M$ and $N$ with respect to midpoints of $AB$ and $BC$. Obviously $OM=OM'$ and $ON=ON'$. $\angle BMN=x$; $\angle BNM=y$. Then $\angle MOM'=180^{\circ}-2x$; $\angle NON'=180^{\circ}-2y$. From this follows, that $\angle M'ON'=2x+2y-180^{\circ}=60^{\circ}$, since $x+y=120^{\circ}$ (you get this from $\triangle BMN)$. Lemma: Let $ABC$ be a triangle with $\angle ABC=60^{\circ}$. A point $O$ lies on the bisector of angle $ABC$, $B$ and $O$ are on different sides of $AC$ and $\angle AOC=60^{\circ}$. Then $O$ is the $B$-excenter of triangle $ABC$. This is not hard to prove. By lemma, $O$ is the $B$-excenter of triangle $BM'N'$. This implies that $M'N'$ is tangent to the incircle

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