XXIII OBM

Consider the following Lemma. Let $ABCD$ be a convex quadrilateral. The lines that connect the midpoints of opposite sides meet in $E$ and the perpendicular bisectors of opposite sides $AB$ and $CD$ meet in $O$. Then the altitudes relative to $AB$ and $CD$ meet in a point $O^{\prime }$ symmetric to $O$ with respect to $E$.



Proof. It's well known that the quadrilateral whose vertices are the midpoints of $ABCD$ is a parallelogram. So $E$ is the midpoint of both its diagonals, in particular $MN$. But since the pairs of lines $OM$, $N{O}^{\prime }$ and $ON$, $M{O}^{\prime }$ are parallel, $M{O}^{\prime }NO$ is also a parallelogram, so the diagonals $MN$ and $O{O}^{\prime }$ meet in their respective midpoints. This means that $E$ is the midpoint of $O{O}^{\prime }$ and we're done.

Let $O_1$ and $O_2$ be the intersection of the perpendicular bisectors of $AB$, $CD$ and $AD$, $BC$, respectively and $O_1^{\prime }$ and $O_2^{\prime }$ be the intersection points of the altitudes relative to $AB$, $CD$ and $AD$, $BC$ respectively. The four altitudes have a common point if and only if $O_1^{\prime }=O_2^{\prime }$, which by the lemma is equivalent to $O_1=O_2$, which is, in turn, equivalent to $ABCD$ being cyclic.