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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.

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  • $\begingroup$ How is O related to the objects of this drawing if it is the center of this circle? $\endgroup$ – miracle173 Feb 16 '17 at 17:48
  • $\begingroup$ @miracle173 no, it is not given that O is the centre.We have to prove that O is the centre with the details given. $\endgroup$ – Navneet Kumar Feb 16 '17 at 17:51
  • $\begingroup$ yes, of course but how is the center of the circle related to the other objects? How can the center of this circe distinguished form other points in the plane? What properties does O have? $\endgroup$ – miracle173 Feb 16 '17 at 17:55
  • $\begingroup$ Hint 1: The distance AO is equal to the distance BO, therefore you draw this double bars. But there are a lot of points with this property. What other property does O have that uniquely determines O. $\endgroup$ – miracle173 Feb 16 '17 at 18:13
  • $\begingroup$ @miracle173 angle(ACB)*2=angle(AOB) $\endgroup$ – Navneet Kumar Feb 16 '17 at 18:20
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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.

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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 $\Delta AOB$.

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    $\begingroup$ I think it is rather complicateto involve the sine for such a prove. Maybe it should be deduced by more elementary concepts and reasoning. $\endgroup$ – miracle173 Feb 16 '17 at 17:51
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    $\begingroup$ It's not that complicated, but I'm happy to see a proof using congruences or something similar. $\endgroup$ – Simon Marynissen Feb 16 '17 at 17:53
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    $\begingroup$ I think I have now a proof using congrences. $\endgroup$ – miracle173 Feb 27 '17 at 10:01
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Case 1: $O \in \triangle(ABC) ^\circ$

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$$|\overline{OA}|=|\overline{OC}|$$ $$\implies \angle(OCA)=\angle(CAO)=:u$$

$$|\overline{OB}|=|\overline{OC}|$$ $$\implies \angle(OCB)=\angle(CBO)=:v$$

$$\angle(AOB)+\angle(BOC)+\angle(COA)=2\pi$$ $$\implies \angle(AOB)\\=2\pi-\angle(BOC)-\angle(COA)\\=2\pi-(\pi-2v)-(\pi-2u)\\=2u+2v\\=2\angle(ACB)$$

Case 2: $O \notin \triangle(ABC)^-$

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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 $O \in \overline{BC}$

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

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