Although
only three folds are needed for the base edges of the tetrahedron,
I’m hoping that more folds are allowed in total. I seem to need three
extra folds, each applied to the original starting shape as follows.

a.Fold so that B coincides with
C. Use the fold to identify the mid-point, M, of BC.
b.Fold so that the edges CA and
CB are collinear and call the crease CD (D on AB).
c.Fold to form a crease through
M while aligning the two parts of crease CD. Call this new
crease ME (E on AC).
d.Find the mid-point, N, of AE by
folding A on to E.
e.Fold along BN,
f.Fold along MN.

The folds along BN, MN and ME can now be used as the edges of the
required tetrahedron ABMN, with A, E and B, C being coincident pairs.

Proof: CM = BM = 2 (from a).
Triangle CEM isosceles (from b & c), so CE = CM =2 therefore CE = BA. EN = AN (from d).
These three underlined statements prove that the edge segments match.