$4+$Given a parallelogram $K L M N$, where $K L=8, K N=3 \sqrt{2}+\sqrt{6}$ and $\angle K N L=45^{\circ}$. On side $K L$, a point $A$ is taken such that $K A: A L=3: 1$. Through point $A$, a line parallel to $L M$ is drawn, on which a point $B$ is chosen inside the parallelogram, and a point $C$ is chosen on side $K N$ such that $K C=A B$. Lines $L C$ and $M B$ intersect at point $D$. Find the angle $\angle L A D$.

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Prove that the lines $A N, B M$, 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

$180^{\circ}-\arccos \frac{\sqrt{6}-\sqrt{2}}{4}=105^{\circ}$.