Make a sketch of this problem. Between △CAD and △CFB, CA=CF(2sides of square ACFG), and CD=CB(2 sides of square BCDE).
And ∠ACD=∠ACB+∠BCD=∠ACB+90° and ∠FCB=∠ACB+∠FCA=∠ACB+90° So ∠ACD=∠FCB Thus △CAD≡△CFB (SAS). Therefore AD=BF. Q.E.D.
Note that AD crosses BF at right angles. It shows that, when △CFB is turned 90 degrees around point C as the center: CF to CA, and CB to CD, △CFB and △CAD fit together.