Solve a tetrahedron with given lengths of three edges and a certain property We are given a tetrahedron $ABCD$ with three known edges: $|AB|=a$, $|BC|=b$ and $|CD|=c$. We also know that all four lines that connect a vertex to the center of the incircle of the opposite face has a mutual point. Does this information define tetrahedron uniquely? If so, how do we find the rest of the edges?
 A: The barycentric coordinates of the incenter are straightforward to find. If $A,B,C$ are three points in the plane and $a=BC, b=AC, c=AB$, the incenter $I$ fulfills
$$ I = \frac{aA+bB+cC}{a+b+c} $$
but the same holds if $A,B,C,D$ are the vertices of a tetrahedron with $AB=a, BC=b, AC=c_1, BD=a_1, CD=c$. In such a case the incenter of $BCD$ lies at
$$ I_{BCD} = \frac{a_1 C + b D + c B}{a_1+b+c} $$
and the incenter of $ABC$ lies at
$$ I_{ABC} = \frac{a C + b A + c_1 B}{a+b+c_1}. $$
The lines $AI_{BCD}$ and $DI_{ABC}$ meet iff the four points
$$ A,\qquad D,\qquad \frac{a_1 C + b D + c B}{a_1+b+c},\qquad  \frac{a C + b A + c_1 B}{a+b+c_1}$$
are coplanar. By saying that through a determinant, iff $ac=a_1 c_1$.
Here it comes the real wonder:

In a tetrahedron $ABCD$, the lines joining each vertex with the
  incenter of the opposite face are concurrent (i.e. there is a sort of 3D-analogue of the Gergonne point) iff the product of the
  lenghts of opposite edges is constant.


In our case, that gives $AD=\frac{ac}{b}$.
