geometry

AB : BC = 3 : 1

Let G  be located on AC  such that  FG is   parallel to   CD

angle GFB  = angle CEB

BF  = BE

And angle CBE  = angle  FBG

So by ASA  triangle FBG  is congruent  to triangle EBC

So  BG  = BC

But since    AF   =  AD   then   GA  also =  GC

But  BC  +  BG   =    GC

And since BG  = BC  then      2BC  =  GC

So 2BC also   =  GA

And  AC =  GC + GA     =    2BC  + 2BC   =  4BC

And AB =   4BC  - BC   = 3BC

So

AB : BC    =   3 : 1            { just as the  guest found !!! }