We use Cartesian coordinate.：） But I use complex number and waste a lot of time.

Let H be the orthocenter of ADC, X be the circumcenter of ADC, then its easy to see that there's a homothety that takes triangle $O_1O_2X$ to triangle $ACH$ since the side are parallel and we can use Desargues theorem The result follows easily

Solution with p_square, BOBTHEGR8, Arwen713 We use DDIT! Let $O_3$ be the circumcenter of $ABC$ and $G_3$ be the centroid. Also, let $M,N$ be the midpoints of $AB,BC$. We also know that in a quadrilateral with perpendicular diagonals $AC, BD$, $AC\cap BD$ has an isogonal conjugate. Let this be $E$. Also, let $AO_2\cap CO_1=F$. Now, applying DDIT on quadrilateral $AMCN$, we get that there's an involution from $O_3$ with pairs $(O_3G_3, O_3B), (O_3M, O_3N)$ and $(O_3A, O_3C)$. But, observe that $O_3O_1M$ are collinear and $O_3O_2N$ are collinear and $O_3,B,E$ are collinear. Now, by DDIT on $AO_2CO_1$, there's an involution swapping $(O_3A,O_3C), (O_3O_1, O_3O_2)$ and $(O_3E, O_3F)$. But $(O_3O_1,O_3O_2)=(O_3M, O_3N)$. So, the two mentioned involutions are the same and as $O_3E=O_3B$, we have $O_3G=O_3F$ but then $F$ lies on $O_3G_3$ i.e. Euler line of $\Delta ABC$. Similarly, we can get the result for $\Delta ADC$.

Let $O_A$ be the circumcenter of $\triangle ABD$ and define $O_B,O_C$ and $O_D$ similarly. Let $H_B$ and $H_D$ be the orthocenters of $\triangle ABC$ and $\triangle ADC$. We will show that quadrilateral $CH_BAH_D$ and $O_AO_BO_CO_D$ are homothetic. Notice that $O_AO_B$ and $CH_B$ are parallel because they are perpendicular to $AB$. This is true for all six sides including the diagonals, and we're done.

Homothety taking triangle \(O_A\)\(O_C\)\(O_D\) to triangle \(CAH_D\).

Let $O_3$ and $O_4$ be the circumcenters of $\triangle ABC$ and $\triangle ADC$ respectively. Also let $H_1$ and $H_2$ be the orthocenters of $\triangle ABC$ and $\triangle ADC$ respectively. It is clear that $O_2O_3\perp BC$ and $AH_1\perp BC \implies AH_1\parallel O_2O_3$ and similarly $AH_2\parallel O_2O_4$. Let $H_1O_3\cap H_2O_4 = F$. The homothety centered at $F$ taking $\triangle FO_3O_4$ to $\triangle FH_1H_2$ will take $O_2$ to $AH_1\cap AH_2 = A$. So $F,O_2,A$ are collinear and similarly $C,O_1,F$ are collinear and thus we are done.

Let $O_B$ and $H_B$ be the circumcenter and orthocenter of $\triangle ABC,$ respectively. Similarly define $O_D$ and $H_D$ for $\triangle ADC.$ Notice $\overline{AC}$ is the radical axis of $(ABC)$ and $(ADC),$ so $\overline{O_BO_D}\perp\overline{AC}$ and $\overline{O_BO_D}\parallel\overline{BD}=\overline{H_BH_D}.$ Also, $\overline{AB}$ is the radical axis of $(ABD)$ and $(ABC),$ so $\overline{O_BO_1}\perp\overline{AB}$ and $\overline{O_BO_1}\parallel\overline{CH_B}.$ Similarly, $\overline{O_CO_1}\parallel\overline{CH_D}$ so $\triangle O_1O_BO_D$ is homothetic to $\triangle CH_BH_D,$ which implies $CO_1$ and the two Euler lines are concurrent. We can analogously show that $AO_2$ passes through this point. $\square$

Let $P,Q$ be the circumcentre and orthocentre of $\triangle ABC$. Then, $AQ\perp BC$ and $PO_2\perp BC$ because $P,O_2$ lie on perpendicular bisector of $BC$. Thus, $AQ\parallel PO_2$ and similarly, $CQ\parallel PO_1$. We also have $O_1O_2\perp BD$ because $BD$ is the radical axis of $(ABD), (CBD)$ and $AC\perp BD$. Therefore $AC\parallel O_1O_2$ so $\triangle QAC$ and $\triangle PO_2O_1$ are homothetic, implying that $PQ$ (the Euler line of $\triangle ABC$), $AO_2$ and $CO_1$ concur. Similarly, the Euler line of $\triangle ADC$ concurs with $AO_2$ and $CO_1$ too.

wow First, we will prove the following lemma, which is the main idea of this solution. Let two lines $\ell_1$ and $\ell_2$ meet at $O$, and consider fixed points $A$ and $C$. Let $\ell$ be a variable line parallel to $AC$, and have $\ell$ intersect $\ell_1$ and $\ell_2$ at $O_1$ and $O_2$ respectively. Then, the intersection of $AO_2$ and $CO_1$, which we call $P$, lies on a fixed line through $O$ as $\ell$ varies. By similar triangles, we have $$P=\frac{A(O_1O_2)+O_2(AC)}{O_1O_2+AC}$$$$P-O=\frac{O_1O_2(A-O)+AC(O_2-O)}{O_1O_2+AC}.$$ Note that both $O_1O_2$ and $O_2-O$ are proportional to the distance from $O$ to $\ell$, but $A-O$ and $AC$ stay constant. Thus, the direction of the vector in the numerator does not change as $\ell$ changes, showing the claim. Let's now assess the situation in the problem. In triangle $\triangle ABC$, there is some line $\ell$ parallel to $AC$ that represents the perpendicular bisector of $BD$. Then, let $\ell_1$ and $\ell_2$ be the perpendicular bisectors of $AB$ and $CB$ respectively. By our lemma, we have that the intersection of $AO_2$ and $CO_1$, which we call $P$, lies on a fixed line. By taking $\ell$ to pass through $O$, we see that this line passes through $O$. Furthermore, by taking $\ell$ to be the $B$-midline of $\triangle ABC$, we see that this line passes through $G$. Thus, $P$ lies on the Euler line of $\triangle ABC$, and similarly it also lies on the Euler lie of $\triangle ADC$, done.

Neat problem! Solved with deduck. Pretty much the same idea as the above solutions, but posting anyways. We denote by $O_b$ , $O_d$ , $H_b$ and $H_d$ respectively, the circumcenters and orthocenters of $\triangle ABC$ and $\triangle ACD$. The following claim is the pith of the problem. Claim : Triangles $\triangle O_1O_2O_b \sim \triangle ACH_b$ and $\triangle O_1O_2O_d \sim \triangle ACH_d$ are pairwise homothetic. Proof : Consider circles $(ABD)$ and $(ABC)$. Since the line joining the centers of two circles is perpendicular to their radical axis, $O_1O_b \perp AB$. Similarly, $O_1O_d \perp AD$ , $O_2O_b \perp BC$ , $O_2O_d \perp CD$ and $O_1O_2 \perp BD$ . Now, by the definition of the orthocenter we also know that $H_bC \perp AB$ , $H_bA \perp BC$ , $H_dA \perp CD$ and $H_dC \perp AD$. Thus, $O_1O_b \parallel CH_b$ , $O_2O_b \parallel AH_b$ , $O_1O_b \parallel AH_d$ , $O_2O_d \parallel CH_d$ and $O_1O_2 \parallel AC$. Thus, triangles $\triangle O_1O_2O_b$ and $\triangle ACH_b$ and triangles $\triangle O_1O_2O_d $ and $ \triangle ACH_d$ have pairwise parallel sides, making these two pairs of triangles homothetic. Now, since it is well known that lines joining corresponding vertices of homothetic triangles are concurrent, $AO_2$ , $CO_1$ , $O_bH_b$ and $O_dH_d$ all concur, as desired. Remarks : This is a really interesting configuration. Let $P$ and $Q$ denote the intersections of the diagonals of quadrilaterals $ABCD$ and $O_1O_bO_2O_d$. Via the same homothety argument it follows that $PQ$ also passes through the aforementioned concurrence point. Further, if $M_Q$ is the Miquel Point of $\triangle ABCD$, $M_Q$ also lies on line $\overline{PQ}$. Further, with some work one can also show that quadrilaterals $(M_QAQO_1)$ and similar are all cyclic.