Problem 1173

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 $O'$ 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 $OM$, $NO'$ and $ON$, $MO'$ are parallel, $MO'NO$ is also a parallelogram, so the diagonals $MN$ and $OO'$ meet in their respective midpoints. This means that $E$ is the midpoint of $OO'$ 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 $O_1'$ and $O_2'$ 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 $O_1' = O_2'$, which by the lemma is equivalent to $O_1 = O_2$, which is, in turn, equivalent to $ABCD$ being cyclic.