Estonian Math Competitions

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

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Alternative solution.

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 $KL$.