Školjka

%V0 Let $ABC$ be a triangle with $AB=AC$ and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D,B$ and $C$ at the point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A,E$ and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incentre of triangle $KAB$. Proposed by Jan Vonk, Belgium and Hojoo Lee, South Korea