In an equilateral trapezoid, the point $O$ is the midpoint of the base $AD$. A circle with a center at a point $O$ and a radius $BO$ is tangent to a straight line $AB$. Let the segment $AC$ intersect this circle at point $K(K \ne C)$, and let $M$ is a point such that $ABCM$ is a parallelogram. The circumscribed circle of a triangle $CMD$ intersects the segment $AC$ at a point $L(L\ne C)$. Prove that $AK=CL$.