Let $ABC$ be a triangle, $(O)$ is the circle through $B$, $C$. $(K)$ is a circle tangent to $AB$, $AC$ and $(O)$. The blue line $(l)$ is the tangent of $(K)$. The circle $(K')$ tangent to $(O)$, $(K')$ tangent to $BC$ at $D$, and $(K')$ tangent to $(l)$ at $E$ (See the figure). How can prove that $D$, $E$, $I$ are collinear.
Converse theorem: Let $ABC$ be a triangle, $(O)$ is the circle through $B$, $C$, $I$ is the incenter of $ABC$. $(K')$ is a circle tangent to $AB$ and $(O)$. $ID$ meets $(K')$ again at $E$. The blue line $(l)$ is the tangent of $(K')$ at $E$. How can prove that there exists a circle tangent to $AB$, $AC$, $(O)$ and the blue line $l$.
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