Question 90
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 $C$ in a 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 incenter of triangle $KAB$.