Question 1050448
<pre><b>
{{{drawing(400,1600/9,-9,9,-1,7,
 
line(-8,0,8,0), line(-8,0,-4,6),line(-4,6,4,6),line(4,6,8,0),
line(-8,0,4,6),line(8,0,-4,6),
locate(-8,0,A), locate(8,0,B),locate(4,6.8,C),locate(-4,6.8,D), locate(-.2,3.9,O), locate(2.23,2.72,E), 
green(line(4,6,2.4,2.8))
 

 )}}} 

Draw CE &#8869; BD.

CE is an altitude of both &#916;DOC and &#916;BOC

The area of &#916;DOC = (1/2)(base)(altitude) = (1/2)*DO*CE
The area of &#916;BOC = (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 &#916;DOC = (1/2)(base)(altitude) = (1/2)DO*CE = (1/2)(2) = 1.

Now extend AC and draw BF &#8869; AC's extension.

{{{drawing(400,1600/9,-9,9,-1,7,
 
line(-8,0,8,0), line(-8,0,-4,6),line(-4,6,4,6),line(4,6,8,0),
line(-8,0,4,6),line(8,0,-4,6),
locate(-8,0,A), locate(8,0,B),locate(4,6.8,C),locate(-4,6.8,D), locate(-.2,3.9,O), locate(5.2,6.6,F), 
green(line(4,6,14,11),line(8,0,4.8,6.4))
 

 )}}} 

Then BF is an altitude of both &#916;BOC and &#916;AOB,

so by the same reasoning as above, the area of
&#916;AOB is twice the area of &#916;BOC.

Therefore &#916;AOB has area 4.

The four triangles &#916;DOC, &#916;AOD, &#916;BOC, and &#916;AOB
make up the entire isosceles trapezoid.

Since &#916;AOD &#8773; &#916;BOC, 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</pre></b>