What a very nice and simple problem. Let $D$ be a point in the circumference such that $DA$ is diameter of the circle and $E$ the point on the perpendicular of $OA$ that lies on the circle. Since $DA$ is a diameter we have $ \angle DKA=90^\circ$ and notice $EO\perp DA$ $\implies $$\angle DOB=90^\circ$ so $DKOB$ is cyclic $\implies$ $\angle ODK= \angle KBC$ Now since $l$ is tangent at $K$ we must have $\angle ADK= \angle AKC$ because they're looking at the same arc (small arc $KA$). So we have $\angle CKB= \angle CBK$$ \implies$ $\triangle CBK$ is isosceles with $KC=BC$. $Q.E.D $ $ \square$.