finiding a ratio in Trapezoid 
Given $ABCD$ a right angle trapezoid, $\measuredangle A=90^{0}$, $AB\parallel DC$.
$O$ is the point of intersection of the diagonals.
$E$ is a point on $BC$ such that: $AB=BE$, $DC=CE$.
$AD=12$, $BC=13$
Need to find the ratio $AO:OC$.

Tried to compute the area and divided it the partial area by the triangles but didn't help.
Any ideas?
Thanks!
 A: The diagram should help.

I didn't really know much about the properties of trapeziums. So a quick search gave this property on wikipedia, and man! It makes the problem damn easy: 

The diagonals cut each other in mutually the same ratio (this ratio is the same as that between the lengths of the parallel sides)

After this, all you have to do is find the sides. The figure tells everything.
A: The given $ABCD$ trapazoid with $\angle BAD = 90^\circ$ is depicted in the following figure with given data:

It is given that $AD=12, \ BC=13$, and $E$ is situated on $BC$ in such a way that $AB=BE$ and $DC=CE$. If you draw a perpendicular line to $CD$ from the point $B$, which meets $CD$ at point $F$, then $AB = DF$. Moreover, the $\triangle BCF$ is a Pythagorian triangle where $BC = 13$ and $BF = 12$, then $FC^2 = 13^2 - 12^2 = 5^2$. Therefore, $FC = 5$.
If $AB= a$, then $DF = a$. Thus,  since $CE = DC$, then $CE = DF + FC =  a + 5$. Similarly, since $BE = AB$, $BE = a$. Therefore, $BC = a + (a + 5) = 13$ and hence, $a= \frac{13-5}{2}= 4$.
As shown in the diagram, using geometric knowledge in parrelel lines, you can show $\triangle ABO$ and $\triangle CDO$ are similar. Hence,
$$\frac{BO}{DO} = \frac{AO}{CO} = \frac{AB}{CD} = \frac{4}{4 + 5} = \frac{4}{9}$$
Consider the right triangle, $\triangle ABD$. Thus, $BD^2 = AB^2 + AD^2= 4^2 + 12^2= 16 \times 10$. Therefore, $BD = 4\sqrt{10}$.
Also, consider the right triangle, $\triangle ACD$. Thus, $AC^2 = DC^2 + AD^2= 9^2 + 12^2= 15^2$. Therefore, $AC = 15$.
Therefore, you can also calculate the distance of $AO, OC, BO,$ and $OD$ easily now.
