Draw CE ⊥ BD.

CE is an altitude of both ΔDOC and ΔBOC

The area of ΔDOC = (1/2)(base)(altitude) = (1/2)*DO*CE
The area of ΔBOC = (1/2)(base)(altitude) = (1/2)*BO*CE = 2 (given)

BO = 2DO because O divides the diagonals in ratio 1:2.

Substitute 2DO for BO in   

(1/2)(2DO)(CE) = 2
DO*CE = 2

The area of ΔDOC = (1/2)(base)(altitude) = (1/2)DO*CE = (1/2)(2) = 1.

Now extend AC and draw BF ⊥ AC's extension.

 

Then BF is an altitude of both ΔBOC and ΔAOB,

so by the same reasoning as above, the area of
ΔAOB is twice the area of ΔBOC.

Therefore ΔAOB has area 4.

The four triangles ΔDOC, ΔAOD, ΔBOC, and ΔAOB
make up the entire isosceles trapezoid.

Since ΔAOD ≅ ΔBOC, they both have area 2, so
adding up the areas of those four triangles, 1+2+2+4, 
we get 9 for the area of isosceles trapezoid ABCD.

Edwin