Area of the shaded region of the intersection between two triangles Given figure below

Find the area of the shaded region.
The only thing I found is DCO triangle is congruent with ABO triangle
$$\frac{AB}{CD}=\frac{AO}{CO}\\
\frac{9}{CD}=\frac{AO}{17}$$
I don't think this lead anywhere. Any clue what to do? Thanks.
 A: One easier way to solve this is to use the Law of Cosines and then apply coordinate geometry.
$\angle DAB = \angle ABC=\theta$, in the equation $9^2+10^2-2(9)(10)\cos(\theta)=17^2$.
Upon solving, you get $\displaystyle \cos\theta=\frac{3}{5}$, which is a very clean angle.
Now, if you set $A=(0,0)$ and $B=(9,0)$ in coordinates, you can find the equations of the line segments $DB$ and $AC$. 
For $AC$, it would be $y = mx$, where $\displaystyle m=\tan(\theta)=\frac{4}{3}$, so $\displaystyle y=\frac{4}{3}x$. 
For $BD$, it would be $y-0=m(x-9)$, in point slope, where $m=-\frac{4}{3}$, this time.
You have both the lines. Now, find the intersection of those lines, take the y-value, and use the base-height formula of a triangle to find the area.
A: Firstly observe that $A,B,C,D$ are concyclic. 
Then find $|CD|$ with using Ptolemy's Theorem. 
Now, you can find $Area(ADCB)$ with using Brahmagupta's formula.
Also, you can find $Area(ACB)=Area(ADB)=\,?$ that will give you an equation related with $Area(DOC)$ and $Area(OAB)$.
With using $\dfrac{Area(DOC)}{Area(OAB)}=\left(\dfrac{|CD|}{9}\right)^2$ , you can find ${Area(OAB)}$
A: I will assume $17$ is from $A$ to the point of intersection of $DB$ and $AC$ which you call $O$. The triangles $ABO$ and $DCO$ are similar so that $AO=BO=x$ and $\angle ADB=\angle ACB=\theta$. Using the law of cosines you get the following equations which you can easily solve for $x$ (without solving for $\theta$)
$$\left(17+x\right)^{2}+100-20\left(17+x\right)\cos\left(\theta\right)=81 \tag{1}$$
$$x^{2}=100+17^{2}-340\cos\left(\theta\right) \tag{2} $$ Once you find $x$ you can use Heron's formula to find the area of $ABO$ so: $$A=\sqrt{\left(x+\frac{9}{2}\right)\left(x-\frac{9}{2}\right)\left(\frac{9}{2}\right)^{2}}=\frac{9}{2}\sqrt{\left(x+\frac{9}{2}\right)\left(x-\frac{9}{2}\right)}$$ $x$ is a nasty number though.
