We consider a simplex $A_0A_1...A_n$ in $n$-dimensional Euclidean $\Bbb E^n.$
Denote by
$Vol(A_0A_2A_3...A_n)=V_1,$
$Vol(A_0A_1A_3...A_n)=V_2,$
$...$
$Vol(A_0A_2A_3...A_{n-1})=V_n$
and
$Vol(A_1A_2A_3...A_n)=V.$
Assume that
$$V_1^2+V_2^2+...+V_n^2=V^2.$$
My question. Are the following set of vectors orthogonal? $$\{\vec{A_0A_1},\vec{A_0A_2},...,\vec{A_0A_n}\}$$
This is false even in three dimensions as can be shown by a simple computation. The philosophical reason is that you are assuming one condition (vanishing of a scalar function on the space of tetrahedra) and hoping to deduce several. There are several converses of pythagoras in three space but they reqire bundling your condition with two other properties of right tetrahedra.
Edit at OP's request. We can assume that the vertices of the tetrahedron are $(0,0,0),(1,0,0),(p,q,0),(r,s,t)$. Then the condition that it be a right tetrahedron is described by the three equations $p=r=s=0$, that of your Pythagoras condition by one $t^2(q^2+(p-1)^2)+((p-1)s-q(r-1))^2=s^2+t^2(p^2+q^2+1)+(ps-qr)^2+q^2$. There are clearly many more tetrahedra that satisfy the latter condition. There are, of course, an infinite number of ways to add two to get a converse. One rather trivial one is to demand that two of the face triangles satisfy the usual Pythagoras. Less trivial: one face satisfies Pythagoras and the volume is $\dfrac 16$ of the product of the appropriate three sides.
