<s>[INST] 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!
 [/INST] 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.
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