Let $\omega$ be a circle with center $O,$ and let $A$ be a point with tangents $AP$ and $AQ$ to the circle. Denote by $K$ the intersection point of $AO$ and $PQ.$ $l_1$ and $l_2$ are two lines passing through $A$ that intersect $\omega.$ Call $B$ the intersection point of $l_1$ with $\omega,$ which is located nearer to $A$ on $l_1.$ Call $C$ the intersection point of $l_2$ with $\omega,$ which is located further to $A$ on $l_2.$ Prove that $\angle PAB = \angle QAC$ if and only if the points $B, K, C$ are on line.