<Qua-2023-718>, hint 

Suppose Lines BA and BD meet at point P and height of triangle PAD be 2h. 

AD=1, BC=4, EF=3

Then heights of triangles PBC=8h, PEF=6h and the area of ADFE=9h-h=8h

area of EFCB=16h-9=7h. 8h/m=7h/n Hence m=8, n=7

Let AD=1, DA'=3, EF=3, FE'=1. AEE'A' is a parallelogram  and so AA'=EE'=4, 4 x 7(m-1)/2=14

Similarly in trapezium BCFE ; ((4 +3) x 6)/2=21.

Hence m+n=15, 14+21=35.