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 
$$(\sqrt{A_1} + \sqrt{A_2})^2$$
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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. 
