Slovenija 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 $K^{\prime }$ and $L^{\prime }$ points $K$, $L$, $K^{\prime }$ and $L^{\prime }$ are concyclic. Since the triangle $BCD$ is acute, $K^{\prime }$ and $L^{\prime }$ lie on the same side of $KL$ and on the same side of $BD$. So, $$\angle D{L}^{\prime }{K}^{\prime }=\angle K{L}^{\prime }{K}^{\prime }=\pi -\angle K^{\prime }LK=\pi -\angle K^{\prime }BD$$ and $B$, $D$, $L^{\prime }$, $K^{\prime }$ are concyclic.