Given a parallelogram $ABCD$, where $AB=5$, $AD=2\sqrt{3}+2$, and $\angle BAD=30^{\circ}$. On side $AB$, a point $K$ is taken such that $AK:KB=4:1$. A line parallel to $AD$ is drawn through point $K$. On this line, inside the parallelogram, a point $L$ is chosen, and on side $AD$, a point $M$ is chosen such that $AM=KL$. The lines $BM$ and $CL$ intersect at point $N$. Find the angle $\angle BKN$.
Prove that the lines $K D, B M$ and $C L$ intersect at one point. For this, use the following statement. Through a point $X$ lying inside a parallelogram, lines parallel to its sides are drawn. Then the two parallelograms formed in this way with the only common vertex $X$ are equal in area if and only if the point $X$ lies on the diagonal of the parallelogram.

## Answer

$75^{\circ}$.