We can make the tetrahedron by 6 diagonal of the attached rectangular parallelepiped.

Attachments:

$ \mbox{Let } O_a, O_b, O_c, O_d \mbox{ center of circumcircle of } \triangle{BCD}, \triangle{ACD}, \triangle{ABC}; \mbox{ and } R_a, R_b, R_c, R_d \mbox{ circumradius this circles.}$ $ \mbox{If O is center of circumscribed sphere and I center of inscribed sphere of tetrahedron } ABCD, \mbox{and } \\ R_a = R_b = R_c = R_d, \mbox{ then: } |OO_a| = |OO_b| = |OO_c| = |OO_d|\Rightarrow O\equiv{I} \Rightarrow \\ \Rightarrow ABCD - \mbox{isoscele tetrahedron}\Rightarrow |AB| = |BD|, |AC| = |BD|, |AD| = |BC|.$ view: Nathan Altshiller-Court: "MODERN PURE SOLID GEOMETRY"(second edition); p.107; 304 Theorem.