BdMO Online Forum

Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$, and let $O$ be the center of its circumcircle. Show that the segments $OA$, $OF$, $OB$, $OD$, $OC$, $OE$ dissect the triangle $ABC$ into three pairs of triangles that have equal areas.

Note that,
$\triangle OAF=\frac{1}{2}\cdot AF\cdot AO\cdot \sin (\angle OAF)=\frac{1}{2}\cdot \cos A\cdot AC\cdot R\cdot \sin (90^{\circ}-\angle C)=\frac{R}{2}\cdot b\cdot \cos C$
Derive similar equations for the other triangles to show that,
$\triangle OAF=\triangle OCD$, $\triangle OBF=\triangle OCE$ and $\triangle OBD=\triangle OEA$.