Denote by $A^{\prime}$ the reflection of $A$ in $B D$. We will show that the quadrilaterals $A^{\prime} B K E$ and $A^{\prime} D L F$ are cyclic, and their circumcircles are tangent to each other at point $A^{\prime}$. From the symmetry about line $B C$ we have $\angle B E K=\angle B A K$, while from the symmetry in $B D$ we have $\angle B A K=\angle B A^{\prime} K$. Hence $\angle B E K=\angle B A^{\prime} K$, which implies that the quadrilateral $A^{\prime} B K E$ is cyclic. Similarly, the quadrilateral $A^{\prime} D L F$ is also cyclic. ! For showing that circles $A^{\prime} B K E$ and $A^{\prime} D L F$ are tangent it suffices to prove that
$$\angle A^{\prime} K B+\angle A^{\prime} L D=\angle B A^{\prime} D.$$
Indeed, by $A K \perp B C$, $A L \perp C D$, and again the symmetry in $B D$ we have
$$\angle A^{\prime} K B+\angle A^{\prime} L D=180^{\circ}-\angle K A^{\prime} L=180^{\circ}-\angle K A L=\angle B C D=\angle B A D=\angle B A^{\prime} D,$$
as required. Comment 1. The key to the solution above is introducing the point $A^{\prime}$; then the angle calculations can be done in many different ways.