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.

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.
