Solve for angle $x$ in a scalene triangle This problem was posted by a friend of mine on Instagram, namely, @dare2solve. Here's the problem:

Given thus scalene triangle, the goal is to solve for the missing angle $x$ in this figure. This problem was actually posted a few months back (in September) on this site, however it was deleted due to the original asker showing no approach of their own.
I'm going to share my own approach as a solution, please let me know if there are any errors in it, and please share your own solutions too. Particularly, is there a way to use Ceva's theorem here to solve for the angles?
 A: The points label are as referenced by OP in his answer to the question.
By triangle angle sum property: $\angle AOC = 80°, \angle AOB = 140° - x, \angle ACB = 50°$
Applying sine law on $\triangle AOB, \triangle AOC, \triangle ABC$, we get:
$$\frac{AB}{\sin{(140°-x)}} = \frac{AO}{\sin{x}} \tag1$$
$$\frac{AC}{\sin{80°}} = \frac{AO}{\sin{30}}\tag2$$
$$\frac{AC}{\sin{20°}} = \frac{AB}{\sin{50°}}\tag3$$
Dividing (2) by (3),
$$\frac{\sin{20°}}{\sin{80°}} = \frac{\sin{50°}AO}{\sin{30°} AB}$$
$$\implies \frac{\sin{50°}}{\sin{30°} AB} = \frac{\sin{20°}}{\sin{80°}AO}  \tag4$$
Multiplying (4) by (1),
$$\frac{\sin{50°}}{\sin{30°}\sin{(140°-x)}} = \frac{\sin{20°}}{\sin{80°}\sin{x}}$$
$$\implies \frac{\sin{50°}}{\sin{30°}\sin{(40°+x)}} = \frac{\sin{20°}}{\sin{80°}\sin{x}}$$
$$\implies \frac{\sin{50°}}{\sin{30°}\sin{(40°+x)}} = \frac{2 \sin{10°} \cos{10°}}{\cos{10°}\sin{x}}$$
$$\implies \frac{\sin{50°}}{\sin{(40°+x)}} = \frac{2 \sin{30°} \sin{10°}}{\sin{x}}$$
$$\implies \frac{\sin{50°}}{\sin{(40°+x)}} = \frac{\sin{10°}}{\sin{x}}$$
From above equation, it can be easily seen that $x = 10°$ satisfies the equation. Moreover, by triangle sum property, we know that ∠ABC is 20°. So, $x$ is less than 20°. The only solution below 20° for the above equation is 10°. Hence, $x = 10°$.
A: Heres my approach to this problem:

1.) Extend segment $BA$ to meet point $O'$ such that $OA=AO'$ and join $O'$ with point $C$. Notice that $\angle OAC$ = $\angle O'AC$ = $70.$
2.) This implies that $\triangle OAC$ and $\triangle O'AC$ are congruent via the SAS property. Therefore, $OC=O'C$ and $\angle ACO$ = $\angle ACO'$ = $30.$ Connect $O$ and $O$ to form an equilateral triangle $\triangle COO'$ where $CO=CO'=OO'$.
3.) Above shows that $\triangle BCO'$ is an isosceles $20-80-80$ triangle, where $BO'=BC.$ Notice that $\angle BO'O$ = $\angle AOO'$ = $\angle BCO$ = $20.$ This implies that $\triangle BOC$ and $\triangle BOO'$ are congruent via the SAS property.
4.) This proves that $\angle O'BO$ = $\angle CBO$ = $x$. Therefore $2x=20$, hence $x=10.$
