If M is the BC vertical line and G is the vertical line, then MG = 1/2 FB (center line), and the triangle CGM is an isosceles right angle, then CM is twice the root number of MG.

It is twice the root of BF, so when BF is 2, CM is the root of 2.

Question (2)

Construction of square CDGH

It is easy to prove that DG = DC, DE = DF and ∠GDE = ∠CDF.

So triangle GDE and triangle DCF are congruent, so ∠DGE is right angle, so e is on GH.

So GE = CF, EH = BF

Take k on CH and make CK = FC, then KH = BF = EH.

So EHK is an isosceles right triangle.

CM is the center line of EK.

So CM = 1/2 EK = root number 2/2 KH =? Root number 2/2 BF

BF/CM = root number 2

Are you sure this is a sophomore problem? . . This auxiliary line is difficult to make.