Let the circle $\omega$ be circumscribed around the triangle $\vartriangle ABC$ with right angle $\angle A$. Tangent to the circle $\omega$ at point $A$ intersects the line $BC$ at point $D$. Point $E$ is symmetric to $A$ with respect to the line $BC$. Let $K$ be the foot of the perpendicular drawn from point $A$ on $BE$, $L$ the midpoint of $AK$. The line $BL$ intersects the circle $\omega$ for the second time at the point $N$. Prove that the line $BD$ is tangent to the circle circumscribed around the triangle $\vartriangle ADM$.