any idea or solution?

Let's call $X_{YZ}$ the projection of X in line YZ, and $X_{MNP}$ the projection of X in plane MNP. Suppose $ AB^2+CD^2 = AC^2+BD^2 = AD^2+BC^2 $. Since $ AB^2-BC^2 = AD^2-CD^2 $, then $B_{AC} = D_{AC}$. This way, using the 3 perpendiculars theorem, we conclude $D_{ABC}$ belongs to line $BB_{AC}$. By the same method, we can conclude $D_{ABC}$ belongs to line $AA_{BC}$, so that $D_{ABC} = H_d$ and $DH_d$ is the height of D in tetrahedron ABCD. Notice that, in plane $DBH_d$, we have $DH_d$ concurs with $BH_b$ [let's call $P$ this concurrency point]. One can prove that, in pairs, the lines of the problem are concurrent. Our goal now is to prove that they are all concurrent at P. Now, P is in plane $DH_dA$ = plane $DH_aA$. Since line $BH_b$ is not in plane $DH_aA$, and since P is the only point in plane $DH_aA$ that is in $BH_b$, then the only possible intersection of $BH_b$ and $AH_a$ is point P. Given that $BH_b$ and $AH_a$ are concurrent, indeed, then they are concurrent at point P. Now we achieved our goal [i will think conversely what happens..]

The converse follows: If $ AH_{a},BH_{b},CH_{c}, DH_{d} $ are concurrent, then $B_{AC}=D_{AC}$, and then: $AB^2-CB^2=AB_{AC}^2-CB_{AC}^2=AD^2 -CD^2$, then $AB^2+CD^2=BC^2+AD^2$ This way, one can prove that $AB^2+CD^2=AC^2+BD^2$, and then it is done. [Here and before i used this fact: in a triangle $ABC$, let $D$ be the foot of height in vertex $A$. Then $AB^2-AC^2=DB^2-DC^2$. This one can be easily proved using Pythagorean Theorem ]