$\triangle ABC$ is equilateral with internal point $O$, find length measure $BO$ $(a)$ This is a problem that was shared to me by my old high school friend from Japan. While I did solve the problem and show him the solution, he remarked that it was a little complicated and he'd prefer a more "algebraic" approach (not sure what that means). This is why, I'm sharing it here. I'll post my own approach as an answer as well!

 A: This is my approach. I'll add an explanation as well!

Here's how I go about it:
1.) Label the triangle $\triangle ABC$ and label the internal point $O$. Now, rotate $\triangle BOC$ by $60$ degrees clockwise such that a new triangle $\triangle ADC$ is formed which is congruent to $\triangle BOC$ with $OC=DC=2a$ and $\angle OCD=60$ (can be proven easily).
2.) Connect points $O$ and $D$ via segment $OD$. Notice that because $\angle OCD=60$, and $OC=DC$, it follows that $\triangle OCD$ is equilateral, therefore $OC=DC=OD=2a$. This also means that $\angle ADO=120-60=60$.
3.) Since $AD=a$ and $OD=2a$, as well as $\angle ADO=60$, we can conclude that $\triangle ADO$ is a triangle that is congruent to a "$30-60-90$ triangle" where the hypotenuse has a measure of $2a$ via the SAS property. Therefore, we can infer that $\triangle ADO$ is a "$30-60-90$ triangle" with $\angle OAD=90$ and $\angle AOD=30$. This implies further that length $OA=a\sqrt{3}=9$. Therefore $a=3\sqrt{3}$.
A: Based on you posted picture and labeling, a very ugly algebraic approach, but without computational effort, could be as below;
assume $B=(0,0)$, then by the law of cosines we have $C=(\sqrt7 a,0)$, and $A=(\frac{\sqrt7a}{2}, \frac{\sqrt21a}{2})$ (we just computed the length of $BC$).
Now, suppose $O=(x,y)$. So we have;
$$x^2+y^2=a^2$$
$$y^2+(\sqrt7a-x)^2=4a^2;$$
hence $7a^2-2\sqrt7ax=3a^2\implies x=\frac{2a}{\sqrt7}$, and $y=\frac{\sqrt3a}{\sqrt7}$. At this stage, it is just needed to compute the length of $OA$;
$$OA^2=9^2=(x-\frac{\sqrt7a}{2})^2+(y-\frac{\sqrt21a}{2})^2\implies$$
$$(\frac{2a}{\sqrt7}-\frac{\sqrt7a}{2})^2+(\frac{\sqrt3a}{\sqrt7}-\frac{\sqrt21a}{2})^2=9^2\implies a=3\sqrt3.$$
