i cant understand the question

The approach that comes to mind is that if you have a tetrahedron in space and inscribe it in a parallelpiped, then those conditions mean that it is actually inscribed in a rectangular prism. Then you would take the limit as height of this prism went to zero. Possibly it would be easier to prove in 3 space (but maybe it is false then...).

My approach is different than the rchlch's It is enough to prove that $ O_1A_1, O_2A_2, O_3A_3$ are concurrent, because of the simetry. Now, the condition $ A_1A_2.A_3A_4 = A_1A_4.A_2A_3$ is equivalent to $ \frac {A_1A_2}{A_2A_3} = \frac {A_1A_4}{A_4A_3}$, from that, the angle bisectors of $ < A_1A_2A_3$, $ < A_1A_4A_3$ and the line $ A_1A_3$ are concurrent, let's say in $ T$. Now let's consider $ S$ the center of the Apolonius circle $ \omega$ of triangle $ A_1A_2A_3$ respect to $ A_2$.Therefore $ B$ is in $ \omega$ iff $ \frac {A_1B}{BA_3} = \frac {A_1A_2}{A_2A_3}$, then it is obvious that $ A_2$ and $ A_4$ belong to $ \omega$. The point $ S$ is the center of the Apolonius circle of the triangle $ A_1A_2A_3$ and $ A_1A_4A_3$, thus, $ S$ belongs to the bisectors of the segments $ A_2S$ and $ A_4S$, i.e. $ S$ is the circumcircle of the triangle $ A_2A_4T$, then $ S$ belongs to the bisector of $ A_2A_4$ as well. But because $ O_1$ and $ O_3$ are the circumcenters of $ A_2A_4A_3$ and $ A_2A_4A_1$, then $ O_1O_3$ is perpendicular to $ A_2A_4$. That means that the intersection of $ O_1O_3$ and $ A_1A_3$ is a center of the Apolonius circle of the triangle $ A_1A_2A_3$. Therefore, the other two intersections between: $ O_1O_2,A_1A_2$ and $ O_2O_3,A_2A_3$ are also center of the Apolonius circles of the triangle $ A_1A_2A_3$. It is well known that these three centers are collinear, then by the Desargues Theorem, the result follows.

orl wrote: Given four points $ A_1, A_2, A_3, A_4$ in the plane, no three collinear, such that \[ A_1A_2 \cdot A_3 A_4 = A_1 A_3 \cdot A_2 A_4 = A_1 A_4 \cdot A_2 A_3, \] denote by $ O_i$ the circumcenter of $ \triangle A_j A_k A_l$ with $ \{i,j,k,l\} = \{1,2,3,4\}.$ Assuming $ \forall i A_i \neq O_i ,$ prove that the four lines $ A_iO_i$ are concurrent or parallel. Or equivalently, prove that the isodynamic points lie on the Neuberg cubic .

I forgot that there're "no three collinear", so a found a configuration that satisfys the problem condicions , the bissector case...

I cannot but recommend to read the solution(s) in the RMC2009 (Romanian Mathematical Competitions 2009), using Apollonius circles followed by Desargues theorem.

Can it be something but a triangle with a point inside? Did someone proved that it can be a quadrilateral?

Yes, it could easily be some kite (even the rhombus) with the equal angles equal to $ \pi/6$.

I thought that I had concluded the problem... I think I've made a mistake... Well, is the triangle with his incenter a valid case?

I have a slightly different approach. Denote by $\omega_{ij}$ the Apollonius circle of points $A_k$ and $A_l$ and ratio $\dfrac{A_iA_k}{A_iA_l}=\dfrac{A_jA_k}{A_jA_l}$, which passes through $A_i$ and $A_j$. It is somewhat well-known that $\omega_{ij}$, $\omega_{il}$ and $\omega_{ik}$ are coaxal and orthogonal to $\odot{A_jA_lA_k}$, so that the radical axis $r_i$ of these three circles passes through $O_i$. Hence $r_i$ is actually the line $A_iO_i$. So $A_iO_i$, $A_jO_j$ and $A_kO_k$ are respectively the radical axes of $\omega_{ij}$ and $\omega_{ik}$, $\omega_{ij}$ and $\omega_{jk}$, $\omega_{ik}$ and $\omega_{jk}$, and therefore concur. It is enough to prove that three of them concur due to symmetry.

Let $ A_1=A, A_2=B, A_3=C, A_4=P, O_1=O_A, O_2=O_B, O_3=O_C$, it suffices to prove $AO_A, BO_B, CO_C$ concur. Notice that $ \angle BPC = \angle BAC + 60$. This is done by taking APA'A'' a cyclic quadrilateral with A'' and A' foots of ange bisectors, and the analogous cyclic quads. And then finish with Trigonometric Ceva

My solution: Lemma (well-known) : Let $ P, Q $ be the isogonal conjugate of $ \triangle ABC $ . Let $ O_P, O_Q $ be the circumcenter of $ \triangle BPC, \triangle BQC $, respectively . Then $ AO_P, AO_Q $ are isogonal conjugate of $ \angle BAC $ . ____________________________________________________________ Back to the main problem: WLOG $ A_4 $ is in $ \triangle A_1A_2A_3 $ . Let $ \triangle N_1N_2N_3 $ be the outer Napoleon triangle of $ \triangle A_1A_2A_3 $ . Let $ F, N $ be the 1st Fermat point, 1st Napoleon point of $ \triangle A_1A_2A_3 $, respectively Since $ A_4 $ is the 1st Isodynamic point of $ \triangle A_1A_2A_3 $ , so $ F $ is the isogonal conjugate of $ A_4 $ WRT $ \triangle A_1A_2A_3 $ . From lemma $ \Longrightarrow A_1O_1, A_2O_2, A_3O_3 $ are concurrent at the isogonal conjugate of $ N $ WRT $ \triangle A_1A_2A_3 $ . Similarly, we can prove $ A_2O_2, A_3O_3, A_4O_4 $ are concurrent $ \Longrightarrow A_1O_1, A_2O_2, A_3O_3, A_4O_4 $ are concurrent . Q.E.D

Assume wlog $A_1$ lies inside triangle $A_2A_3A_4$, it is clear that it is the first isodynamic point of this triangle. Clearly, triangle $O_2O_3O_4$ is orthologic with triangle $A_2A_3A_4$ and the centers of orthology are obviously, $O_1$ and $A_1$. Thus, showing that $A_iO_i$ for $i=2,3,4$ concur automatically establishes that the concurrency point lies on $A_1O_1$ by the Sondat's Theorem. For the concurrency, use the "Distance Ceva" as follows: Let $\angle A_i=\theta_i$ for $i=2,3,4$. Then, we want \begin{align*} \Pi_{i =2,3,4} \left(\frac{d(O_i,A_iA_{i+1})}{d(O_i,A_iA_{i+2})}\right)=1 \end{align*}where $A_5=A_2$ and $A_6=A_3$. This holds since the above mentioned product in fact equals the quantity \begin{align*} \Pi_{i=2,3,4} \left(\frac{\sin (\theta_{i+2}-60^{\circ})}{\sin (\theta_{i+1}+60^{\circ})} \right) \end{align*}which is obviously equal to $1$. The result follows.

Note that $\frac{A_2A_4}{A_3A_4}=\frac{A_1A_2}{A_1A_3}$, so $A_4$ lies on the $A_1$-Appolonius circle of $A_1A_2A_3$. Similarly, it lies on the other two Appolonius circles as well. It is well-known that the Appolonius circles of a triangle are coaxial and that its circumcenter lies on its common radical axis. Hence, the four lines we want to prove the concurrency of are the radical axes of the Appolonius circles of the four points taken three at a time. Call these radical axes $\ell_1,\ell_2,\ell_3,\ell_4$, where $\ell_i=A_iO_i$. Call the Apollonius circles $C_{12},C_{13},C_{14},C_{23},C_{24},C_{34}$, where $C_{ij}$ is the Appolonius circle of $A_i$ and $A_j$ that passes through the other two points. Let $\ell_1$ intersect $\ell_2$ at $P$. Then, $p(P,C_{23})=p(P,C_{34})=p(P,C_{24})$ and $p(P,C_{13})=p(P,C_{34})=p(P,C_{14})$, so $p(P,C_{24})=p(P,C_{34})=p(P,C_{14})$ and $p(P,C_{23})=p(P,C_{34})=p(P,C_{13})$, which respectively mean that $P$ lies on $\ell_3$ and $\ell_4$, as desired.

Note that $\frac{A_2A_4}{A_3A_4}=\frac{A_1A_2}{A_1A_3}$, so $A_4$ lies on the $A_1$-Apollonius circle of $A_1A_2A_3$. Similarly, it lies on the other two Apollonius circles as well. Also since the Apollonius circles of a triangle are coaxial and are orthogonal to the circumcircle, so the radical axis passes through the circumcenter. By radical axis theorem on these Apollonius circles twice, we get that $O_1A_1, O_2A_2, O_3A_3$ are concurrent. Now by similarly repeating this argument or by Sondat's theorem on $O_1O_2O_3$ and $A_1A_2A_3$ we get our result. EDIT: realised that this is almost the same as the previous solution.

It is enough to prove lines $A_1O_1$, $A_2O_2$, $A_3O_3$ are concurrent or parallel. Rename the points to get the following statement: Quote: Let $P$ be a point in the plane of $\triangle ABC$ (not lying on any sideline of $\triangle ABC$) such that \[AP \cdot BC = BP \cdot CA = CP \cdot AB.\]Let $O_A$ denote the circumcenter of $\triangle BPC$; define $O_B$ and $O_C$ similarly. Prove that lines $AO_A$, $BO_B$, $CO_C$ are concurrent. Let $A'$ denote the inverse of $A$ in $(BPC)$; define $B'$ and $C'$ similarly. Claim. $BCB'C'$ is cyclic. Proof. Note $\tfrac{B'A}{B'P} = \tfrac{BA}{BP} = \tfrac{CA}{CP} = \tfrac{C'A}{C'P}$ and so the four points lie on an Appolonius circle of $\overline{AP}$. $\square$ Likewise $CAC'A'$ and $ABA'B'$ are cyclic. It is easy to see that $A$, $B$, $C$, $A'$, $B'$, $C'$ are not all concyclic (note $\measuredangle BA'C = \measuredangle BAC \pm 120^{\circ}$), so lines $AA'$, $BB'$, $CC'$ are concurrent or parallel; i.e. lines $AO_A$, $BO_B$, $CO_C$ are concurrent or parallel.

I am unable to find a possible configuration for the parallel case. Does anyone elaborate on it?

It suffices to prove that $A_2O_2$, $A_3O_3$ and $A_4O_4$ are concurrent. Let $B_i$ be the image of $A_i$ under the inversion with center $A_1$ and radius $1$ for $i\in\{2,3,4\}$. Since $B_2B_3=\frac{A_2A_3}{A_1A_2\cdot A_1A_3}$, $B_2B_3B_4$ is equilateral. Under our inversion, $O_2$ goes to the point symmetric to $A_1$ wrt $B_3B_4$, so $A_2O_2$ goes to the circle through $A_1$ and $B_2$ with center on $B_2B_3$. It suffices to prove that three centers of such circles are collinear. If $P_2$, $P_3$, $P_4$ are circumcenters of $A_1B_3B_4$, $A_1B_2B_4$ and $A_1B_2B_3$, this is equivalent to $P_2P_3P_4$ being perspective to $B_2B_3B_4$. However, $B_iP_i$ concur at the center of $B_2B_3B_4$, so we're done by Desargues's theorem.