Let $ABCD$ a tetrahedron and $E,F,G,H,I,J$ on $AB,BC,CA,CD,DA,DB$ s.t. $AE\cdot EB=BF\cdot FC=CH\cdot HD=DI\cdot IA=AG\cdot GC=DJ\cdot JB$. Show that $E,F,G,H,I$ lies on a sphere.
I know that is sufficient to prove that the perpendicular in the circumcircle of $(EFG)$ and the perpendicular in the circumcircle of $(GHI)$ are concurent.
Some introductory words to get closer to the solution first. The points $E,F,G,H,I,J$ are on the sides of the tetrahedron. Let us consider their reflections $E',F',G',H',I',J'$ w.r.t. the mid points of the corresponding sides. Then we have for instance $EA\cdot EB=E'B\cdot E'A=E'A\cdot E'B$. In the given generality, replacing $E$ by $E'$ would also fit the given data in the problem. So these points are also (a posteriori) on the sphere $\Sigma$ to be found. Since $E,E';F,F';G,G'$ are on $\Sigma$, the intersection of $\Sigma$ with the plane of the six points is a circle, and its center is on the side bisectors of $EE'$, of $FF'$, of $GG'$. Which are the side bisectors of $AB$, of $BC$, of $CA$. So the center of this circle $(EE'FF'GG')$ is the circumcenter of $(ABC)$. So the center of $\Sigma$ (if it exists) is the center of the sphere $(ABCD)$. Oh, we can now start to work... We forget everything, and start over again.
Let $O$ be the center of the sphere $(ABCD)$.
We work first only in the plane $(ABC)$. Let $\odot(ABC)$ be the circumcenter of the triangle $\Delta ABC$. Let $R$ be its radius, $O_{ABC}$ its center. Then the power of $E$ w.r.t. to this circle is $EA\cdot EB$, it is the difference of $R^2$ and the squared distance between $E$ and $O_{ABC}$.
The problem gives three equal powers for $E,F,G$ w.r.t. $\odot(ABC)$, so they are on a circle centered in $O_{ABC}$. In particular, $E,F,G$ have equal distance w.r.t. $O$.
Use the other triangles of the given tetrahedron to conclude. (Using OP's observation, the perpendicular in the circumcircle of $\odot(EFG)$ and the perpendicular in the circumcircle of $\odot(GHI)$ are concurent in a point, yes, in $O$.)