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 $\triangle LOC = \triangle BOC = \triangle BOK$, 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.