A tetrahedron $A-BCD$,and let $$S_{ABC}=a,S_{ACD}=b,S_{ABD}=c,S_{BCD}=d$$ and such $$d^2=a^2+b^2+c^2$$
show that: then tetrahedron $A-BCD$ is Trirectangular Tetrahedron.
I know this reslut,if $A-BCD$ is Trirectangular Tetrahedron.then we have $$d^2=a^2+b^2+c^2$$ and I can prove this. But the inverse problem,I can't see it,and can't prove it.I hope someone can know? Thank you
I'm going to change your notation slightly, for better symmetry: $$a = |\triangle BCD| \qquad b = |\triangle CDA| \qquad c = |\triangle DAB| \qquad d = |\triangle ABC|$$ (So, "$x$" is the area of the face opposite vertex "$X$".)
Consider this tetrahedral Law of Cosines for Faces: $$a^2 = b^2 + c^2 + d^2 - 2 c d \cos \angle AB - 2 d b \cos\angle AC - 2 b c \cos \angle AD$$ where "$\angle XY$" indicates the dihedral angle along edge $XY$.
Certainly, if the tetrahedron is tri-rectangular (what I call "right-corner") with hypotenuse-face $\triangle BCD$, then the cosines vanish and we have de Gua's Theorem, $$a^2 = b^2 + c^2 + d^2 \qquad (\star)$$
However, the converse isn't true. Assuming $(\star)$, the Law of Cosines for Faces reduces to $$0 = c d \cos \angle AB + d b \cos\angle AC + b c \cos \angle AD$$ This relation can hold without the cosines simultaneously vanishing. (Contrast this with Law of Cosines for Triangles, which has just a single cosine term.) In fact, if you fix hypotenuse-face $\triangle BCD$, then $(\star)$ is (usually) satisfied by a continuum of tetrahedra with vertex $A$ on some ellipsoid.
