TR-MO 2007 Problem 1

In an acute triangle $ABC$ , the circle with diameter $AC$ intersects $AB$ and $AC$ at $K$ and $L$ different from $A$ and $C$ respectively. The circumcircle of $ABC$ intersects the line $CK$ at the point $F$ different from $C$ and the line $AL$ at the point $D$ different from $A$ . A point $E$ is choosen on the smaller arc of $AC$ of the circumcircle of $ABC$ . Let $N$ be the intersection of the lines $BE$ and $AC$ . If $AF^{2}+BD^{2}+CE^{2}=AE^{2}+CD^{2}+BF^{2}$ prove that $\angle KNB= \angle BNL$ .