If ABCDEFGH is a cube measured as being 1 unit as volume, then corresponding line segments measure:
1 = AB = BC = CD = DA = AF = BG = CH = DE = EF = FG = GH = HE,
2^.5 = ca. 1.414 = AC = BF = DF = DH = HF, and
3^.5 = ca. 1.732 = CF, so,
Smallest 2 volumes each with faces of 3 right isoceles triangles [BGF, BGH and FGH, and DEF, DEH and EFH] and 1 equilateral [BFH and DFH]; 3 mutually perpendicular unit 1 leg measures and 3 hypotenuses of ca. 1.414, BFGH and DFHE are congruent;
Mid-sized 2 volumes each with faces of 2 right isoceles triangles [ABC and ABF, and ADC and ADF] and 2 triangles of unit, square and cube diagonal sides [BCF and ACF, and ACF and DCF]; ADCF and ABCF are reflective; and
Largest 2 volumes each with faces of 1 right isoceles triangle [BCH and DCH], 1 equilateral triangle [BFH and DFH] and 2 of unit, square diagonal and cube diagonal sides [BCF and CFH, and CFH and DCF]; DCHF and BCHF are reflective.
Triangles ACF and HCF measure sides 1 unit, 1 square diagonal ca. 1.414, and one cube diagonal ca. 1.732.
Is this a dissection of a cube into six congruent tetrahedra, or a pair of congruent tetrahedra triplets?
Tetrahedron: (Lengths) 1; 2^.5; 3^.5
FCBA: AB, AF, DE; AC, BF; CF
FCAD: AD, DC, DE; AC, DF; CF
FCDE: CD, DE, EF; CE, DF; CF
FCEH: CH, EF, EH; CE, FH; CF
FCHG: CH, FG, GH; CG, FH; CF
FCGB: BC, BG, FG; BF, CG; CF
As EF, HF, GF, BF, AF and DF sequentially alternate about CF, respectively, each corresponding tetrahedron; through a common non-symetrical triangle; reflects its complementary tetrahedron:
.. FCBA; ACF;
FCGB; BCF; ..*
So, because the observed triangular non-symetries noted above preclude mirror image congruence, alternate tetrahedra,
FCBA, FCDE and FCHG are congruent
FCAD, FCEH and FCGB are, also.
*.. an "ellipset" pair connects a loop.
Pending further review, see