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.