If the faces of tetrahedron $ABCD$ are congruent triangles, then $AB = CD$, $AC = BD$, $AD = BC$ $ \triangle BCD \cong \triangle ADC \cong \triangle DAB \cong \triangle CBA $ sum of the face angles at $A$ : $ \angle BAC + \angle CAD + \angle DAB = \angle CDB + \angle DBC + \angle BCD = \pi $ similarly $ \angle ABC + \angle CBD + \angle DBA = \pi $, $ \angle ACB + \angle BCD + \angle DCA = \pi $, $ \angle ADB + \angle BDC + \angle CDA = \pi $ Conversely, if the sum of the face angles at each vertex of tetrahedron $ABCD$ is $\pi$ $ \angle BAC + \angle CAD + \angle DAB = \pi $, $ \angle ABC + \angle CBD + \angle DBA = \pi $, $ \angle ACB + \angle BCD + \angle DCA = \pi $, $ \angle ADB + \angle BDC + \angle CDA = \pi $ 1. by trihedral angle theorem, $ \angle CAD + \angle DAB > \angle BAC $, then we have $ \angle BAC < \pi/2 $ this holds for all 12 angles above, that is, they are acute angles. Note if $\theta$ and $\psi$ are acute angles, then $\theta \ge \psi$ if and only if sin $\theta \ge $ sin $\psi$ 2. denote the circumradius of $ \triangle BCD $ as $R_A$, the circumradius of $ \triangle ADC $ as $R_B$, the circumradius of $ \triangle DAB $ as $R_C$, the circumradius of $ \triangle CBA $ as $R_D$ without loss of generality, assume $R_A$ is the largest by law of sines $2R_A$ sin $ \angle BDC = BC = 2R_D $ sin $ \angle BAC $, and since $ R_A \ge R_D $, then $ \angle BAC \ge \angle BDC $ and similarly we can deduce $ \angle CAD \ge \angle CBD $ $ \angle DAB \ge \angle DCB $ adding above we get $ \pi \ge \pi $ so all inequality are equality, $R_A = R_B = R_C = R_D$ and $ \angle BAC = \angle BDC $, $ \angle CAD = \angle CBD $, $ \angle DAB = \angle DCB $, by law of sines again $2R_B$ sin $ \angle DCA = DA = 2R_C $ sin $ \angle DBA $, and since $ R_B = R_C $, then $ \angle DCA = \angle DBA $, and similarly $ \angle ACB = \angle ADB $, $ \angle ABC = \angle ADC $ 3. now we know the 12 angles are in 6 pairs, then since all 12 angles add to $4\pi$, picking one angle in each pair: $ \angle BAC + \angle CAD + \angle DAB + \angle DBA + \angle ADB + \angle ABC = 2\pi $ substract $ \angle BAC + \angle CAD + \angle DAB = \pi $, we have $ \angle DBA + \angle ADB + \angle ABC = \pi $ compare to angles of $\triangle ABD$ results $ \angle ABC = \angle BAD $ then $ AC = 2R_D$ sin $ \angle ABC = 2R_C $ sin $ \angle BAD = BD $ similarly we can deduce $AB=CD$, $AD=BC$ so $ \triangle BCD \cong \triangle ADC \cong \triangle DAB \cong \triangle CBA $ (SSS)

The if part is obvious the only if part: make the faces on a plane,that is on plane ABC let D1,D2,D3 be the point that BCD1 congruent to BCA D2,D3 the same way Then D2D3A are collinear so ABC is the midpoint triangle of D1D2D3 then we get the result