, 2021

By symmetry it suffices to show that $AO_2$, $CO_1$ and the Euler line of $\triangle ABC$ are concurrent. Note that $O_1O_2 \parallel AC$. Let $\ell_A, \ell_C$ be the perpendicular bisector of $BC, AB$, respectively. Also denote by $M_A, M_C$ the midpoints of $BC, AB$, respectively. Let $G, O, H$ be the centroid, circumcenter, orthocenter of $\triangle ABC$ and let $\ell$ be its Euler line. Then
$$\begin{aligned} (G, O; H, \ell \cap AO_2) &= (AG, AO; AH, AO_2) = (AM_A, AO; A\infty_{\ell_A}, AO_2) \\ &= (M_A, O; \infty_{\ell_A}, O_2). \end{aligned}$$
Similarly $(G, O; H, \ell \cap CO_1) = (M_C, O; \infty_{\ell_C}, O_1)$. Now since $M_A M_C \parallel O_1 O_2 \parallel AC$, we know (by perspectivity at $\infty_{AC}$) that $(M_A, O; \infty_{\ell_A}, O_2) = (M_C, O; \infty_{\ell_C}, O_1)$. Therefore $(G, O; H, \ell \cap AO_2) = (G, O; H, \ell \cap CO_1)$, showing that $\ell, AO_2, CO_1$ are concurrent, as desired.

Remark. The cross ratio here can be replaced by Menelaus theorem applied for multiple times. Cross ratios are used just to make proof concise.

Alternative proof. Again, we have $O_1O_2 \parallel AC$ and it suffices to show that $AO_2, CO_1$ and the Euler line of $\triangle ABC$ are concurrent. Let $M_1, M_2$ be the midpoints of $AB, BC$, respectively. Then note that $M_1M_2 \parallel O_1O_2 \parallel CA$, and so by Desargues theorem we know that $M_1O_1 \cap M_2O_2, CO_1 \cap AO_2$ and $CM_1 \cap AM_2$ are collinear. Note that $M_1O_1 \cap M_2O_2$ is the circumcenter of $\triangle ABC$, and $CM_1 \cap AM_2$ is the centroid of $\triangle ABC$. Therefore $CO_1 \cap AO_2$ lies on the Euler line of $\triangle ABC$. $\square$