Very easy by barycetric coordinates because $X$ is first isodynamic point of triangle $ABC$. By Ceva some calculates (not hard).

Nguyenhuyhoang wrote: Let $X$ be a point inside triangle $ABC$ such that $XA.BC=XB.AC=XC.AC$. Let $I_1, I_2, I_3$ be the incenters of $XBC, XCA, XAB$. Prove that $AI_1, BI_2, CI_3$ are concurrent.

Let $(I)$ be incircle of triangle $ABC$ $D=XI_{1} \cap BC \Rightarrow AD$ is the bisector $\widehat{BAC} $ . Let $X_{1} = AI_{1} \cap XI$ So, $\frac{X_{1}I}{X_{1}X}.\frac{AD}{AI}.\frac{I_{1}X}{I_{1}D}=1.$ Similar for $X_{2}, X_{3}$. Easy get $\frac{X_{1}I}{X_{1}X}= \frac{X_{2}I}{X_{2}X}= \frac{{X_3}I}{X_{3}X}$ So $X_{1} \equiv X_{2} \equiv X_{3} $ Q.E.D

Looks projective! This solution was found in collaboration with pi37. His observation lead to this finish. We note that $X$ is the Appolonius point, and in this instance would be the first. Suppose $BI_1 \cap XC = K$ and $\triangle A'B'C'$ be the cevian triangle of $I$ wrt $\triangle ABC$ were $I$ is the incentre. We have $\dfrac{XK}{KC} = \dfrac{XB}{BC} = \dfrac{XA}{AC}$, hence $K$ is also the foot of the angle bisector from $A$ in $\triangle AXC$, meaning $K \in AI_2$. So, $\triangle XI_1I_2$ and $\triangle ABC$ are perspective. Note that $XI_2\cap BC = A'$ by Appolonius circle. Hence, if we let $I_C = I_1I_2 \cap AB$, then $I_CA'B'$ are collinear. Indeed, this means $I_C$ is the harmonic conjugate of $C'$ wrt $AB$, so $I_C$ is the intersection of the exterior angle bisector at $C$ with $AB$. If we define $I_A, I_B$ similarly, then we note that $I_A, I_B, I_C$ are collinear (desagues on $\triangle A'B'C'$ and $\triangle ABC$ if in doubt). Hence $\triangle I_1I_2I_3$ is perspective with $\triangle ABC$, so done.

Let $XI_2 \cap AC=E$ and $XI_3 \cap AB=F$, since $X$ is the intersection of 3 Apollonius circles, we have $F,E$ are the foot of the angle bisector of $\angle C$ and $\angle B$ respectively. Let $D$ be the foot of the outer angle bisector of $\angle A$, using Menelaus it's easy to show that $EF$ passes through $D$. Then use Desargues theorem for $\triangle I_1I_2I_3$ and $\triangle ABC$, we are done.

Another solution. Let $SABC$ be a tetrahedron such that: \[SA\cdot BC=SB\cdot AC=SC\cdot AB.\] If points $A'$, $B'$, $C'$ are incenters of triangles $SBC$, $SAC$ and $SAB$ respectively, then by the condition lines $AA'$, $BB'$ and $CC'$ concur at one point. Now if Appolonius spheres pairs $(A,B)$, $(B,C)$ and $(C,A)$ intersect at circle $\omega$, we can tending $S$ to $X$ along the $\omega$ and we have configuration in our problem.

A lot of good stuff going on in this thread! It's really nice to see that. Here's mine: We know that it suffices to prove that $\triangle ABC$ and $\triangle I_1I_2I_3$ are perspective. Let $BI_3$ meet $AS$ at $D$, hence $\frac {AD}{DS}=\frac {AB}{BS}=\frac {AC}{CS}$, which means that $D$ is on $CI_2$ $\Rightarrow$ $BI_3,CI_2$ intersect on $AS$. Now similarily define $E,F$. Hence $\triangle DEF$ and $\triangle ABC$ are perspective. By Desarges' theorem on $BEI_3$ and $CFI_2$ we have $I_2I_3$ goes through the intersection of $EF,BC$ and similar for the others. Hence $\triangle ABC$ and $\triangle I_1I_2I_3$ are perspective and we are done. Q.E.D

Burii wrote: Another solution. Let $SABC$ be a tetrahedron such that: \[SA\cdot BC=SB\cdot AC=SC\cdot AB.\] If points $A'$, $B'$, $C'$ are incenters of triangles $SBC$, $SAC$ and $SAB$ respectively, then by the condition lines $AA'$, $BB'$ and $CC'$ concur at one point. Now if Appolonius spheres pairs $(A,B)$, $(B,C)$ and $(C,A)$ intersect at circle $\omega$, we can tending $S$ to $X$ along the $\omega$ and we have configuration in our problem. Beauty Though, this proof for some reason seems like a 3d shorter version of my 2d proof ?

IDMasterz wrote: Burii wrote: Another solution. Let $SABC$ be a tetrahedron such that: \[SA\cdot BC=SB\cdot AC=SC\cdot AB.\] If points $A'$, $B'$, $C'$ are incenters of triangles $SBC$, $SAC$ and $SAB$ respectively, then by the condition lines $AA'$, $BB'$ and $CC'$ concur at one point. Now if Appolonius spheres pairs $(A,B)$, $(B,C)$ and $(C,A)$ intersect at circle $\omega$, we can tending $S$ to $X$ along the $\omega$ and we have configuration in our problem. Beauty Though, this proof for some reason seems like a 3d shorter version of my 2d proof ? It seems that yes:D

Let $AI_1$ meet $XI$ at $T$. Note that $$\frac{XT}{TI}=(XI, T\infty) \overset{A}{=} \frac{\left(\frac{XI_1}{I_1A_1}\right)}{\left(\frac{IA}{AA_1}\right)}=\left(\frac{BX+XC}{BA+AC} \right) \left(\frac{AB+BC+CA}{BC}\right)$$and it is clear that this is symmetric in $A, B, C$, so $T$ is the concurrency point. $\blacksquare$ Remark: The hard part about this solution was to make the gutsy claim that these lines meet on $XI$. On that note, I made a conjecture that $B, C, I_2, I_3$ are concyclic (see why this trivialises the problem), but made no further attempts towards it. I'd be glad to see if it's true or not