Because DE‖BC, EF‖AB
So the angle ADE=EFC, the angle C=AED, and the quadrilateral BFED is a parallelogram.
So △ADE is similar to △EFC, and BF=DE.
Because s △ ade: s △ EFC = 1: 4.
So DE:FC= 1:2.
So BF: FC = 1: 2.
Because triangle DFB is as high as triangle EFC.
Then: sdfb: s△ EFC = BF: FC = 1: 2.
So DFB=2,
So the area of quadrilateral BFED =2S DFB=4.