Clearly, A will remain as (0,0);f1 will make B as (0,4),f2 will make it (12,4) and f3 will make it (4,8);f1 will make C as (2,4),f2 will make it (14,4) and f3 will make it (5,9). Finally, f1 willmake D as (2,0),f2 will make it (2,0) and f3 will make it (1,1). So, finally we get A≡(0,0),B≡(4,8), C≡(5,9) and D≡(1,1). Therefore, mAB=