All-Soviet-Union Mathematical Olympiad

Solution:

Let $\Gamma$ be the circumcircle of $DEF$. Let $OD$, $OE$, $OF$ meet it again at $A''$, $B''$, $C''$ respectively. Then the figure $O'A'B'C'$ must be similar to $OA''B''C''$. So to prove that $OD$ is parallel to $O'A'$, we have to prove that $AO$ is perpendicular to $B''C''$.

So $AO$ meets $B''C''$ at $D''$. Now since $\square AFC'' = 90^{0}$ and $\square AD''C'' = 90^{0}$, both $F$ and $D'$ lie on the circle diameter $AC''$. Hence $AO - OD'' = OF - OC''$. Similarly, $BO$ meets $C''A''$ at $E''$, and $CO$ meets $A''B''$ at $F''$, and $BO - OE'' = OD - OA''$ and $CO - OF'' = OE - OB''$. Hence $OD'' \cdot OA = OE'' \cdot OB = OF'' \cdot OC$. So using the similarity, $OD' - OA = OE' - OB = OF' - OC$.