Demonstrate the following .... The diagonals of a trapezoid (ABCD where AD parallel to BC ) divide it into four triangles all having one vertice in O. Knowing that the area of triangle $\triangle AOD$ is $A_1$ and of $\triangle BOC$ is $A_2$, demonstrate that the area of the trapezoid is 
$$(\sqrt{A_1} + \sqrt{A_2})^2$$
Thank you very much if you are now looking at my problem!
 A: Let $AD= a$ (smaller side). $BC=b$. Height of trapezoid $=h$.
Then,
$ABC+BDC-A_2+A_1=ABD+ACD-A_1+A_2=\text{Area of trapezoid}$
Let height of $AOD$ = y , then height of $BOC=h-y$
$A_1=ay/2$.
$A_2=b(h-y)/2$
Find area of other triangles using $base*height/2$
Find y.
Find $(\sqrt{A_1}+\sqrt{A_2})^2$
Simplify to well known formula of area of trapezoid.
A: Note that $\triangle ABD$ and $\triangle ADC$ have the same area (same base, equal heights). They have $\triangle AOD$ in common, so the areas of the two parts where they differ are equal, That is, $\triangle AOB$ has the same area as $\triangle DOC$.  
Call the area of each by the name $x$. 
Look at $\triangle ABC$. It is divided into two parts, with bases $AO$ and $OC$, and the same heights. Thus
$$\frac{x}{A_2}=\frac{OA}{OC}.\tag{1}$$
A similar calculation using $\triangle ACD$ shows that
$$\frac{x}{A_1}=\frac{OC}{OA}.\tag{2}$$
Multiply the left sides of (1) and (2), and the right sides. We get
$$\frac{x^2}{A_1A_2}=1,$$
which tells us that $x=\sqrt{A_1A_2}$ and we are finished. 
