How to prove that O is the centre of the circle?? 
In the image,It is given that OA=OB and angle(AOB)=2*angle(ACB).
Then,prove that O is the centre of the circle.
I need some hint for solving it.Please help.
 A: By the law of sines, you can deduce that the diameter of the circle is $\frac{AB}{sin(x)}$, hence if you can prove that $AO$ or $BO$ is equal to $\frac{AB}{2sin(x)}$, then $O$ is the center of the circle. You can prove this by using the law of cosines in the triangle $\Delta AOB$.
A: If it is given that ACB is the circle through points A,C,B, then suppose the center is some point N different from O.  Join NA and NB.  Therefore $\angle ANB = 2\angle ACB$ [Euclid III, 20].  But $\angle AOB = 2\angle ACB$.  Therefore $\angle ANB = \angle AOB$. Join NO and extend it to P on the circle.  Then if O is within $\angle ANB$ and P is on arc AB, $\angle AOP$ exterior to $\triangle AON$ is greater than the opposite interior $\angle ANO$ [Euclid I, 16]. Likewise exterior $\angle BOP$ is greater than interior $\angle BNO$. Therefore the whole $\angle AOB$ is greater than the whole $\angle ANB$, contrary to the supposition. Likewise we get a contradiction if N lies within $\angle AOB$, or if NA or NB intersects OB or OA, respectively.  Therefore O is the center of circle ACB.        
A: Case 1:  $O \in \triangle(ABC) ^\circ$

$$|\overline{OA}|=|\overline{OC}|$$
$$\implies \angle(OCA)=\angle(CAO)=:u$$
$$|\overline{OB}|=|\overline{OC}|$$
$$\implies \angle(OCB)=\angle(CBO)=:v$$
$$\angle(AOB)+\angle(BOC)+\angle(COA)=2\pi$$
$$\implies \angle(AOB)\\=2\pi-\angle(BOC)-\angle(COA)\\=2\pi-(\pi-2v)-(\pi-2u)\\=2u+2v\\=2\angle(ACB)$$
Case 2:  $O \notin \triangle(ABC)^-$

We have
$$\angle(AOB)=2\pi-\angle(BOD)-\angle(COA)=2v=2\angle(ACB)$$
Here we used Thales' Theorem (cf case 3).
Case 3:  $O \in \overline{AC}$ or  $O \in \overline{BC}$

$$\angle(AOB)=\pi-\angle(COA)=2\angle(AOC)$$
