Given a parallelogram $ABCD$, where $AB=3$, $AD=\sqrt{3}+1$, and $\angle BAD=60^{\circ}$. On side $AB$, a point $K$ is taken such that $AK:KB=2:1$. 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 $X$ inside a parallelogram, lines parallel to its sides are drawn. Then the two parallelograms formed with the only common vertex $X$ are equal in area if and only if the point $X$ lies on the diagonal of the parallelogram.

## Solution

We will use the following statement. Through a point $X$ inside a parallelogram, lines parallel to its sides are drawn. Then the two parallelograms formed with the only common vertex $X$ are equal in area if and only if the point $X$ lies on the diagonal of the parallelogram. Let the lines $B M$ and $K D$ intersect at point $N$. We will prove that the line $C L$ passes through point $N$. For this, draw lines through point $N$ parallel to the sides of the original parallelogram. Let the line parallel to $A B$ intersect sides $B C$ and $A D$ at points $P$ and $Q$, respectively, and the second line intersect sides $A B$ and $C D$ at points $R$ and $S$, respectively. Extend $L M$ and $K L$ to intersect sides $B C$ and $C D$ at points $F$ and $G$, respectively. Let segments $M F$ and $R S$ intersect at point $E$, and segments $K G$ and $P Q$ intersect at point $H$. Since point $N$ lies on the diagonal $B M$ of parallelogram $A B F M$ and on the diagonal $K D$ of parallelogram $A K G D$, we have $S_{\text{ARNQ}} = S_{\mathrm{NPFE}}$ and $S_{\mathrm{ARNQ}} = S_{\mathrm{NHGS}}$. Therefore, $S_{\mathrm{NPFE}} = S_{\mathrm{NHGS}}$, so parallelograms $H P F L$ and $E L G S$ are equal in area. Consequently, point $L$ lies on the diagonal $C N$ of parallelogram $N P C S$. Therefore, the line $C L$ passes through point $N$. From triangle $A K D$, using the cosine theorem, we find that

$$
\begin{gathered}
K D^{2}=A K^{2}+A D^{2}-2 \cdot A K \cdot A D \cdot \cos 60^{\circ}=2^{2}+(\sqrt{3}+1)^{2}-2 \cdot 2 \cdot(\sqrt{3}+1) \cdot \frac{1}{2}=6 \\
\cos \angle A K D=\frac{K D^{2}+K A^{2}-A D^{2}}{2 \cdot K D \cdot K A}=\frac{4+6-(\sqrt{3}+1)^{2}}{2 \cdot 2 \cdot \sqrt{6}}=\frac{\sqrt{6}-\sqrt{2}}{4}
\end{gathered}
$$

Therefore,

$$
\angle A K D=\arccos \frac{\sqrt{6}-\sqrt{2}}{4}=75^{\circ}, \angle B K N=180^{\circ}-\angle A K D=180^{\circ}-75^{\circ}=105^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-49.jpg?height=663&width=1038&top_left_y=772&top_left_x=516)

## Answer

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