Let $AD,\ BE,\ CF$ be, the altitudes of the given triangle $\triangle ABC$ with circumcircle $(O)$ and orthocenter $H.$ Through $H$ we draw the lines parallel to $AC,\ AB,$ which intersect $AB,\ AC,$ at points $P,\ Q$ respectively and it is enough to prove that $OM\perp PQ,$ where $O$ is the circumcenter of $\triangle ABC$ and $M\equiv AH\cap PQ.$ Let be the point $T\equiv (M)\cap (O),$ where $(M)$ is the circumcircle of $AEHF$ and we have that the line segments $AT,\ EF,\ BC,$ are concurrent at one point so be it $S,$ as the radical center of the circles $(M),\ (O),\ (K),$ where $(K)$ is the circumcircle of $BCEF.$ So, from the harmonic pencil $A.SBDC$ $($ from the complete quadrilateral $AEHFBC$ $),$ and from $PM = MQ$ we conclude that $PQ\parallel AT\equiv AS$ $,(1)$ Because of $(1)$ and $OM\perp AT$, we conclude that $OM\perp PQ$ and the proof is completed. Kostas Vittas.

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Dear Mathlinkers, this nice problem which I sax on Mathlink, but where, is an application of the basic one: A' the midpoint of ABC. Then H bissect the "perpendiculat to A'H at H. For a nice proof we can use the butterfly theorem. Sincerely Jean-Louis

I have a nice proof for this problem: Let $R$ be the midpoint of $AB$, $S$ be the midpoint of $AC$, then $OR \perp AB$ and $OS \perp AC$, so $OMRP$ and $OMQS$ are cyclic. Note that $RM \parallel BH \parallel OS$ and $SM \parallel CH \parallel OR$. Thus $ORMS$ is a parallelogram, so $\triangle ORM$ and $\triangle OSM$ have congruent circumcircles. Then $OP$ and $OQ$, diameters of circles $OMRP$ and $OMQS$ respectively, are equal, so $MP=MQ$.

Dear MLs This problem is Theorem 11 in Darij's brilliant paper on Butterfly "On cyclic quadrilaterals and the butterfly theorem". Still, there is a strong connection with Archimedes Prop. #2 from Lemmas book. Make your own choice. M.T.