69th Belarusian Mathematical Olympiad

The angle $ADC$ of the parallelogram $ABCD$ equals $40^{\circ }$. The point $K$ is given such that the segments $AK$ and $BC$ intersect, $AK=BC$ and $\angle BAK=80^{\circ }$. The point $L$ is given such that the segments $CL$ and $AD$ intersect, $CL=AB$ and $\angle BCL=80^{\circ }$. Find the angles of the triangle $BKL$.