Problem

Source: Turkey TST 2000 P5

Tags: geometry, rhombus, circumcircle, trigonometry, angle bisector, geometry proposed



Points $M,\ N,\ K,\ L$ are taken on the sides $AB,\ BC,\ CD,\ DA$ of a rhombus $ABCD,$ respectively, in such a way that $MN\parallel LK$ and the distance between $MN$ and $KL$ is equal to the height of $ABCD.$ Show that the circumcircles of the triangles $ALM$ and $NCK$ intersect each other, while those of $LDK$ and $MBN$ do not.