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I read in a book that the center of the outcircle(intersect with each vertex) of a triangle is the point of intersection of the perpendicular bisector of $AB$ and the circle that every point on it makes the angle $2C$ with $A$ and $B$.But it has two intersections But we have only one single out circle.So why do we have two centers?

enter image description here

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    "But it has two intersections": who, what, where ??? "why do we have two centers": which two ? – 2017-02-15
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    no chance of two centers what so ever – 2017-02-15
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    Can you provide a picture about your question? It is not clear what you are really asking. – 2017-02-15
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    @Seyed Picture aded – 2017-02-15
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    @YvesDaoust Picture aded. – 2017-02-15
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    @TahaAkbari: you missed a fragment of the original text. "is the point of intersection of the perpendicular bisector of AB and ... " – 2017-02-15
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    Yves is right: the proper claim is not only that the center of the circumcircle lies on the bisector of $AB$, but *also* that it lies on the bisectors of $BC$ and $CA$ as well. In short: your memory was good, but not perfect. The second part of your memory is a slightly muddle statement of the inscribed angle theorem (https://en.wikipedia.org/wiki/Inscribed_angle#Theorem). – 2017-02-15

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First, could you present the construction of circle that every point on it makes the angle $2C$ with $A$ and $B$ ?

If we assume, that this circle have a radius $r>0$, then:

  • If this circle has points on both sides of $AB$, then there is point $C'$ on this circle that $AC'B=0^{\circ}$ or $AC'B=180^{\circ}$ (two points, where the circle intersects with line $AB$). But for the triangle with right angle $ACB$ (then $AC'B=180^{\circ}$) the circle must lie fully on the line $AB$, so it must be a single point ($r=0$)

  • On the other hand, if the circle lies fully on one side of $AB$, there exists a poinf $C''$ on it, that $AC''$ and $BC''$ intersects with the circle in points $A'\neq C''$ and $B'\neq C''$. Then for every point D on arc $A'B'$ we have $\angle AC''B > \angle ADB$.

So this circle can not have a positive radius. Thus it must be a single point. Or some different type of curve.

Second, we know two things:

  • if $O$ is the center of outer circle of triangle $ABC$, then the angle $\angle AOB$ is central angle related with inscribed angle $\angle ACB$, so $\angle AOB=2 \angle ACB$

  • the centre of the outer circle lies exactly on the intersection of bisectors of each side. In particular it lies somewhere on bisector of side $AB$.

See, that the sentence In particular it lies somewhere on bisector of side $AB$ doesn't mean that every point on this bisector is the centre of the outer circle. On the other words, there is only one way implication (not equivalence!): $$O \text{ is the centre} \Rightarrow O \text{ lies on bisector}$$

I belive your 'theorem' was

If $O$ is the center, then it satisfies these two conditions

instead of

$O$ is the center, if it satisfies these two conditions

or even

$O$ is the center, if and only if it satisfies these two conditions

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    For someone whose first language is apparently not English, you've done a beautiful job of explaining something that is tangled up with the subtleties of English usage. :) – 2017-02-15
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    @JohnHughes Thank you :) – 2017-02-15
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Here is a correct figure.

enter image description here

It is quite remarkable that if you move a vertex alongthe circle, the angle beteen the incident sides doesn't change (except if it passes another vertex).

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The centre of the outcircle does indeed lie on the perpendicular bisector of AB, and it makes an angle of $2C$ with $A$ and $B$ (i.e. $A\hat XB=2C$). But the points with the second property do not form a circle; they form a circular arc. If $C<90^{\circ}$ then only the part of your circle above the line $AB$ is relevant. The second point you've marked, which lies below $AB$, actually gives an angle of $A\hat XB=180^{\circ}+2C$, so it is not the centre. If $C>90^{\circ}$ then it is the part of the circle below $AB$ which matters.