50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010)

Let the diagonals of the quadrilateral $ABCD$ intersect at $O$. Altitudes of the triangles $BCD$ and $ACD$ lie on $AC$, so their orthocenters $K$ and $M$ too. Analogously, points $L$, $Q$ lie on $BC$ (Fig.06).

Note that $BK\parallel CL$, because $BK\perp AD$ and $CL\perp AD$. It follows that $\angle BKC=\angle KCL=\angle ACL=90^{\circ }-\angle CAD=90^{\circ }-\angle DBC=90^{\circ }-\angle OBC=\angle BCK$.

Fig.06

Now it's easy to see that $△LOC=△BOC=△BOK$, which implies $OC=OK$ and $BO=OL$. Since $CK\perp CL$ it follows that the quadrilateral $LCBD$ is a rhombus. Analogously, the quadrilateral $ADMQ$ is a rhombus too. Finally, we have that the quadrilateral $KLMQ$ is an image of $ABCD$ according to the central symmetry relative to the point $O$, which implies that the quadrilaterals $KLMQ$ and $ABCD$ are equal.