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I am considering the problem of finding the center of a given circle C of radius r. I currently know 3 elementary ways (ruler+compass construction) of doing this : - choose 2 points on the circle, draw the bisector, and draw another bisector involving a third point on the circle - choose 2 points on the circle, draw the bisector, get a diameter then find the middle point - choose 2 points on the circle, draw a chord, then a perpendicular line and get a diameter (Thales theorem).

I was looking for other elementary method (not involving trigonometry/analytic geometry). Then I heard something quite confusing for me : with two points on the circle C, call them A and B and "close enough", draw the circle C1 of center A and radius AB, and the circle C2 of center B and radius AB, they met at D('inner' wrt the circle C) and E('outer' wrt the circle C). Call F the other intersection of the circles C and C1, then draw (FD) which intersects the circle C at G; it seems that the circle of center G and radius r contains the three points: B, D and the center of the circle C.

I have no idea if this is possible to prove this property using only elementary things. Any hint please? Comments are welcome (and please excuse my not-so-good english :/ )

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    The circle centered on $G$ certainly has the stated property, but you haven’t actually found the center of the original circle yet. – 2017-01-04
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    I think I’d try working backwards from the original circle and the one centered at $G$ and try to prove that $D$, $F$ and $G$ are colinear. – 2017-01-04
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    Seems interesting, but this needs to define G without the line, I don't see how to do it. – 2017-01-04
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    Working backwards, you assume the radius of the given circle is known. Draw a circle of the same radius centered at $B$ and choose one of the intersections as $G$. Note that $G$ depends *only* on the choice of $B$, which may be a clue to producing a proof. – 2017-01-04
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    @amd: Yes: The straight $DE$ and the circle centered at $G$ meets in the center of $C$. (And in $D$) – 2017-01-05
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    @zoli Of course, but the problem as stated doesn’t mention any such line. – 2017-01-05
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    @MathieuDity: Have you taken a look at the solution I gave? Is there any further question? – 2017-01-06

2 Answers 2

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Based on the described construction and using the OP's notations, we have the following figure:

![enter image description here

It is enough to demonstrate that $DG=r$, the radius of the circle.

By the construction, the triangle $ABD$ is equilateral and then $AB=BD=AF$. $BDG$ is an isosceles triangle because the triangles $ABD$ and $ADG$ are congruent. $\angle AGD = \angle BGD$ because / and $AB=AD=AF$ (See the Wiki article called Inscribed angle)

Now, connecting $A$ and $F$ with the center $O$ we get the green triangle $AFO$. To show that $DG=FO=r$, it is enough to prove that the triangles $BDG$ and $AFO$ are congruent.

We have already seen that the bases of these triangles equal. What remains is to show that $\angle AOF = \angle BGD$.

This is easy. Just look at the figure and you will see that the $\angle BDG =2\times\angle AGD = \angle AOF$. (See the Wiki article called Inscribed angle.)

Finally, this is how to use the construction at stake to find the center of the circle:

enter image description here

  1. Construct $E$, $F$, and $D$ and then $G$ as described.
  2. The straight $ED$ and the circle centered at $G$ through $D$ will meet in $O$.

Note

This construction is not that bad compared to the most frequently used one in the case of which you construct two bisecting perpendiculars. In the case of this present construction it is enough to draw the line through $F$ and $D$ which already exist after constructing the first bisecting perpendicular.

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Filling in on the main answer. It takes an argument to show that the triangle BGD is isosceles. We can't apply any usual triangle congruence theorem that I can see.

Here is one approach: take $A, B$, and $D$ as above. Construct a line through $A$ which bisects $BD$. Let the point where it crosses the circle be $G'$. Note the construction forces angle $AG'B$ to equal angle $AG'D$. Draw a line thru $G'$ and $D$ and call the point where this intersects the circle $F'$. The arc $AF'$ equals the arc $BA$ because angle $AG'B$ equals angle $AG'F'$. But this forces $F'=F$. But then $G'$ is the point where the circle meets the line through $F$ and $D$. So $G'=G$.

Perhaps there is a simpler way.