(I)     AB * BC  =  DAD
         *    :       *
(II)    EC :  F  =    F
      -----------------
(III)  BCD *  G  = BBEA
       (IV)  (V)    (VI)


Solution:
(II)    F >= 4
(II.1)  F != 5   =>  F != 0 (mod 5)
(II.1)  F * F  ==  C  (mod 5)
(V.1)   F * G  ==  C  (mod 5)
Therefore F*F == F*G (mod 5). As F != 0 (mod 5) we can cancel F
so we get F == G (mod 5).
(IV)    A * E <= B  =>  E < B
So the product (II) EC is larger than the product (V) BC.
So G = F + 5. As G <= 9 we get F <= 4.
Therefore F = 4 and G = 9.
(II)    C = 6 and E = 1.
(V)     B = 3
(I.1)   D = 8
(III.1) A = 2
 

        23 * 36  =  828
         *    :       *
        16 :  4  =    4
      -----------------
       368 *  9  = 3312


References:

Jürgen Köller;
  Cryptogramme,
  www.mathematische-basteleien.de