We prove that $AKDL$ is a parallelogram; this implies the desired claim since $AD$ and $KL$ are diagonals of this quadrilateral. We prove at first that $KD \parallel AL$.

If $K$ lies between $A$ and $B$ (Fig. 19) then by inscribed angles $\angle BAC = \angle BCD$ and $\angle BCD = \angle BKD$. Consequently, $\angle BAL = \angle BAC = \angle BKD$, implying $KD \parallel AL$.

If $B$ lies between $A$ and $K$ (Fig. 20) then by inscribed angles $\angle BAC = \angle BCD$ and $\angle BCD = 180^\circ - \angle BKD$. Consequently, $\angle BAL + \angle BKD = \angle BAC + \angle BKD = 180^\circ$, implying $KD \parallel AL$.

If $A$ lies between $K$ and $B$ (Fig. 21) then by inscribed angles $\angle BAC = 180^\circ - \angle BCD$ and $\angle BCD = \angle BKD$. Consequently, $\angle BAL = 180^\circ - \angle BAC = \angle BKD$, implying $KD \parallel AL$ again.

Analogously, we can show that $LD \parallel AK$. Altogether, this establishes that $AKDL$ is a parallelogram.

Fig. 19

Fig. 20

Fig. 21

By inscribed angles, $\angle ABC = \angle ALK$, implying that the triangles $ABC$ and $ALK$ are similar. Let the lines $AD$ and $KL$ intersect at point $N$.

Fig. 19

Let $M$ be the midpoint of the side $BC$. It is known that the symmedian line drawn through a vertex of a triangle and the lines tangent to the circumcircle of the triangle at the other two vertices meet in one point; hence $N$ is the point of intersection of the symmedian line drawn through vertex $A$ of the triangle $ABC$ with line $KL$. By the definition of symmedian, $\angle BAM = \angle CAN = \angle LAN$, whence $M$ and $N$ are corresponding points in similar triangles $ABC$ and $ALK$. Thus $N$ bisects the line segment $KL$.