We only need to call on Big Pete Pythagoras:
E 6-b B
F 3-b A 3 C
My points A and B are same as yours; the "fold line" is ED.
I'm leaving left side of diagram out: making its length b, so EB = EA = 6-b;
plus length of AF becomes 3-b.
Let CD = a; then DB = DA = 4-a
Right triangle AEF: 4^2 + (3-b)^2 = (6-b)^2 : leads to b = 11/6
Right triangle ACD: 3^2 + a^2 = (4-a)^2 : leads to a = 7/8
We can now use right triangle ADE (or BDE) to get fold line ED; using ADE:
AE = 6-b = 25/6
AD = 4-a = 25/8
ED = SQRT[(25/6)^2 + (25/8)^2] = 125/24 : 5.208333333......