Mongolian Mathematical Olympiad

Let $H$ be intersection point of the line $D{O}_1$ with circumcircle of the triangle $ABC$. Assume that $C{O}_2\cap EH=O_2^{\prime }$. Now let's prove that $O_2\equiv O_2^{\prime }$.

Let $\angle MAC=\angle DAC=\varphi$, $\angle CBM=\angle CBE=\psi$. Since points $A$, $C$, $D$, $E$ are concyclic, $$\begin{array}{rl}\angle O_{1}HE& =\angle DHE=\angle DHC+\angle CHE=\\ & =\angle DAC+\angle CBE=\varphi +\psi .\end{array}(\ast )$$

It is easy to show that $$\angle M{O}_{2}C=2\angle MAC=2\varphi ,\angle C{O}_{1}M=2\angle CBM=2\psi .$$ Also from $O_{2}M=O_{2}C$ and $O_{1}M=O_{1}C$ follows $$\angle O_1{O}_{2}C=\frac{1}{2}\angle M{O}_{2}C=\varphi \text{ and }\angle C{O}_1{O}_2=\frac{1}{2}\angle C{O}_{1}M=\psi .(\star \star )$$

Therefore $\angle O_{1}C{O}_2=180^{\circ }-\varphi -\psi$. (★★★) Combining (*) and (★★★), we get $$\angle O_{1}C{O}_2^{\prime }=\angle O_{1}C{O}_2=180^{\circ }-\varphi -\psi =\angle O_{1}HE=\angle O_{1}H{O}_2^{\prime }.$$ Thus we conclude that points $O_1$, $C$, $O_2^{\prime }$, $H$ are concyclic. By (★★), $\angle O_1{O}_2^{\prime }C=\angle O_{1}HC=\angle DHC=\angle DAC=\varphi =\angle O_1{O}_{2}C$. Hence we have $O_2^{\prime }\equiv O_2$ that shows the line $E{O}_2$ passes through the point $H$. Analogously, it is possible to prove that the point $H$ lies on the line $F{O}_3$.