BF=10cm, but CF varies depending on the length of BC. CF=BF-BC, and this can be seen diagrammatically by drawing a line parallel to the parallelogram's base to meet the sides AD and BC at D' and C'. The parallelogram ABC'D' replaces ABCD. BF remains at 10cm, but CF is not the same length as C'F and D'C' is intercepted at E'.
However, angles BAF=DAF=CEF=C'E'F=x=(1/2)DAB, where DAB is the bisected angle.
Angle BCD=BC'D'=DAB=2x (opposite internal angles of parallelogram).
Angle AED=AE'D'=BAF=x (alternate angles on parallel lines) and BAF=DAE=D'AE'=x, so the triangles AED and AE'D' are isosceles. BCD and BC'D' are the external angles of triangles ECF and E'C'F respectively, so angle BFA=x because BCD and BC'D'=2x=CEF+BFA=C'E'F+BFA=x+BFA. Therefore triangle ABF is isosceles and AB=BF=10cm. Clearly CF is not the same length as C'F, so it remains indeterminate. BF is fixed at 10cm.
Corollary
Since BF=10 then CF must be less than 10cm, otherwise construction would not be possible.
Let's assume CF>0. Since BC+CF=BF, CF=BF-BC=10-BC>0 and CF<10, and so 0<BC<10.
If, for example, BC=5 then CF=5.
Since the angle DAB can be any value, we can make the problem simpler geometrically by considering it to be 90 degrees making the parallelogram into a rectangle. (If the rectangle becomes a square or the parallelogram becomes a rhombus, then AF becomes a diagonal and CF is zero.) If AD=BC=5 then E will bisect DC because DAE=DAF=BAE=BAF=AED=45, and AD=DE=BC=EC=5. So, using the special case of a rectangle, or a square, it can easily be seen that CF has no constraints to define it other than the range 0<CF<10cm.
