11-th Czech-Slovak-Polish Match, 2011

Let $P$ be the midpoint of $BD$ and $Q$ be the common point of lines $BD$, $KM$ and $LN$. Without losing generality assume that point $B$ lies between $Q$ and $D$. By Tales theorem, $PM\parallel AB$ and $PN\parallel CD$. Therefore, $\angle PNL=\angle NLC=\angle MKA=\angle KMP$. Note that this implies points $Q,M,P,N$ to be concyclic, as $\angle QNP+\angle QMP=180^{\circ }$ and points $M,N$ lie on different sides of the line $BD$. Therefore, $\angle KMN=\angle QMN=\angle QPN=\angle BDC$. Moreover, $\angle LNM=180^{\circ }-\angle QNM=180^{\circ }-\angle QPM=\angle MPD=\angle ABD$. $\square$