Slovenian Mathematical Olympiad, 2008

We use the Law of sines for the triangle $BAL$,
$$\frac{\sin(\angle BAL)}{|BL|} = \frac{\sin(\angle ALB)}{|AB|},$$
and for the triangle $CAL$,
$$\frac{\sin(\angle LAC)}{|CL|} = \frac{\sin(\angle CLA)}{|AC|}.$$
Since $\angle ALB = \pi - \angle CLA$, we have $\sin \angle ALB = \sin \angle CLA$. From the two equations above and $\angle BAL = \angle LAC$ we get
$$\frac{|AB|}{|AC|} = \frac{|BL|}{|CL|}.$$
Similarly, $\frac{|AD|}{|AC|} = \frac{|DK|}{|CK|}$.
(In any triangle the bisector of an angle divides the opposite side in the ratio equal to the ratio of the lengths of the other two sides. This is a well-known fact and the proof was not required. It was sufficient to note that since $AL$ is the bisector of $\angle BAC$, we have $\frac{|AB|}{|AC|} = \frac{|BL|}{|CL|}$.)
Since $|AB| = |AD|$, these two equalities imply $\frac{|BL|}{|CL|} = \frac{|DK|}{|CK|}$. So,
$$\frac{|CB|}{|CL|} = \frac{|BL| + |CL|}{|CL|} = \frac{|BL|}{|CK|} + 1 = \frac{|DK|}{|CK|} + 1 = \frac{|DK| + |CK|}{|CK|} = \frac{|CD|}{|CK|}.$$
Triangles $DBC$ and $KLC$ are similar (they share the angle at $C$ and $\frac{|CB|}{|CL|} = \frac{|CD|}{|CK|}$). Thus, the line $KL$ is parallel to the diagonal $BD$. Because of the right angles at $K'$ and $L'$ points $K$, $L$, $K'$ and $L'$ are concyclic. Since the triangle $BCD$ is acute, $K'$ and $L'$ lie on the same side of $KL$ and on the same side of $BD$. So,
$$\angle DL'K' = \angle KL'K' = \pi - \angle K'LK = \pi - \angle K'BD$$
and $B$, $D$, $L'$, $K'$ are concyclic.