Balkan Mathematical Olympiad

From the orthogonal triangle $ACD$ we have: $\angle ECL=90^{\circ }-\hat{A}$. From the inscribed quadrilateral $ELCM$ we have: $\angle EML=\angle ECL=90^{\circ }-\hat{A}$. Also $\widehat{EMC}=90^{\circ }$ (since $EC$ is a diameter of the circle $(c_1)$). Hence we have: $$\angle LMC=90^{\circ }-\angle EML=90^{\circ }-(90^{\circ }-\hat{A})=\hat{A}.(1)$$ From equality (1) we conclude that the quadrilateral $ALMB$ is cyclic and let $(c_2)$ be its circumcircle. Also the quadrilateral $NDKC$ is cyclic, because $ND\hat{C}=NK\hat{C}=90^{\circ }$. Let now $(c_3)$ be the circumcircle of the quadrilateral $NDKC$. We observe that: The line $LM$ is the radical axis of the circles $(c_1)$ and $(c_2)$. The line $KC$ is the radical axis of the circles $(c)$ and $(c_1)$. The line $AB$ is the radical axis of the circles $(c)$ and $(c_2)$. Hence the lines $LM$, $KC$ and $AB$, are concurrent, say at $P$ (radical center of the three circles $(c)$, $(c_1)$ and $(c_2)$). Therefore we have $$PK\cdot PC=PL\cdot PM(2)$$ Also from circle $(c_3)$ we have the equality: $$PD\cdot PN=PK\cdot PC(3)$$ Finally from (2) and (3) we get the equality $PD\cdot PN=PL\cdot PM$ from which we conclude that the quadrilateral $DLMN$ is cyclic.