Q:

Triangles formed by diagonals of trapezoids

$\Delta$ AOB and $\Delta$ DOC should be equal in area. Correct me if I am wrong.
Given: Trapezoid ABCD with ratio $\frac{area \Delta AOB}{area\Delta ABD}$ = $\frac{3}{4}$.
I am trying to find (1) Ratio of Area $\Delta$AOD to $\Delta$BOC; (2) Ratio of Area $\Delta$COD to Trapezoid ABCD
I know that the final solution (1) is 1:9 and (2) is 3:16, but I do not understand how to get to these answers.
Please explain how to get to the answer, and if possible, how might I find different ratios like $\Delta$BDC to trapezoid ABCD, or $\Delta$BDC to $\Delta$ABD?
Note: I have also seen this post but I'm not sure it's relevant to what I'm asking.

A:

You can use the result of that post to conclude that the ratio is 
$$\frac{a^2}{b^2}$$
The ratio is also 
$$\frac{\frac{1}{2} a h_1}{\frac{1}{2} b h_2}$$
where $h_1,h_2$ are heights of $\Delta AOD$ and $\Delta BOC$, respectively. Now since 
$$\frac{\text{area} \Delta AOB}{\text{area} \Delta ABD}=\frac{3}{4}$$
$$\frac{h_1}{h_2}=\frac{1}{3}$$
which gives you
$$\frac{a^2}{b^2}=\frac{a}{b} \cdot \frac{1}{3}$$
So $\frac{a}{b}=1/3$, that means the ratio of area of $\Delta AOD$ and $\Delta BOC$ is $1:9$. 
I think you can then find the other one using similar argument.

