APMO

Let $M$ and $N$ be midpoints of sides $BC$ and $AC$, respectively. Notice that $\angle MOC=\frac{1}{2}\angle BOC=\angle EAB$, $\angle OMC=90^{\circ }=\angle AEB$, so triangles $OMC$ and $AEB$ are similar and we get $\frac{OM}{AE}=\frac{OC}{AB}$. For triangles $ONA$ and $BDA$ we also have $\frac{ON}{BD}=\frac{OA}{BA}$. Then $\frac{OM}{AE}=\frac{ON}{BD}$ or $BD\cdot OM=AE\cdot ON$.

Denote by $S(\Phi )$ the area of the figure $\Phi$. So, we see that $S(OBD)=\frac{1}{2}BD\cdot OM=\frac{1}{2}AE\cdot ON=S(OAE)$. Analogously, $S(OCD)=S(OAF)$ and $S(OCE)=S(OBF)$.

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Alternative solution.

Let $R$ be the circumradius of triangle $ABC$, and as usual write $A$, $B$, $C$ for angles $\angle CAB$, $\angle ABC$, $\angle BCA$ respectively, and $a$, $b$, $c$ for sides $BC$, $CA$, $AB$ respectively. Then the area of triangle $OCD$ is $$S(OCD)=\frac{1}{2}\cdot OC\cdot CD\cdot \sin (\angle OCD)=\frac{1}{2}R\cdot CD\cdot \sin (\angle OCD)$$ Now $CD=b\cos C$, and $$\angle OCD=\frac{180^{\circ }-2A}{2}=90^{\circ }-A$$ (since triangle $OBC$ is isosceles, and $\angle BOC=2A$). So $$S(OCD)=\frac{1}{2}Rb\cos C\sin (90^{\circ }-A)=\frac{1}{2}Rb\cos C\cos A$$ A similar calculation gives $$\begin{array}{rl}S(OAF)& =\frac{1}{2}OA\cdot AF\cdot \sin (\angle OAF)\\ & =\frac{1}{2}R\cdot (b\cos A)\sin (90^{\circ }-C)\\ & =\frac{1}{2}Rb\cos A\cos C,\end{array}$$ so $OCD$ and $OAF$ have the same area. In the same way we find that $OBD$ and $OAE$ have the same area, as do $OCE$ and $OBF$.