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enter image description here If I'm given vertices of a convex polygon (in the attached image, they are D,E,F,G and H) if we know that inside the polygon there exists a point (say O) for which each angle created by any two adjacent two vertices and the O are equal. That means, angle DOE, angle DOH, angle HOG, angle GOF and angle FOE all are equal. How to find O?

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    $\begingroup$ The point $O$ does not necessarily exist. Under the assumption that the angle $\theta:=\angle DOE$ is constructible or given, then you can simply create the circle passing through $D$ and $E$ such that the arc $DE$ supports the angle $\theta$. Do the same with $E$ and $F$, $F$ and $G$, etc. If all these circles meet at a single point, then it is your required point $O$. If not, then $O$ does not exist. The point $O$ is guaranteed to exist, if the given polygon is a triangle or it is regular. $\endgroup$ – Batominovski Jul 15 '15 at 20:15
  • $\begingroup$ For example, a rectangle (that is not a square) does not have such $O$. But then, the question is, given such $O$ exists, how to find it? $\endgroup$ – peterwhy Jul 15 '15 at 20:30
  • $\begingroup$ If there is an even number of vertices you can draw 2 diagonals, they meet at $O$. The case of an uneven number of vertices seems more difficult $\endgroup$ – Ward Beullens Jul 15 '15 at 20:34
  • $\begingroup$ I'm assuming there exists such an $O$, like in the question. $\endgroup$ – Ward Beullens Jul 15 '15 at 20:37
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For a given $n$-gon $DEF\ldots$ and given that such $O$ exist, then all of the angles $\angle DOE=\angle EOF=\ldots = \frac{2\pi}n$ can be found. Denote the double of that angle as $\theta = 2\angle DOE$.

For any two neighbouring edges, say $DE$ and $EF$, draw an isosceles triangle for each edge with the edge as the base side, and $\theta$ as the opposite angle. The new isosceles triangle should lie inside the $n$-gon. Call the two new opposite vertices $A$ and $B$ respectively.

At $A$ with radius $AE$, and at $B$ with radius $BE$, draw two circles. The two circles intersect at $O$ and at $E$.

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    $\begingroup$ It's not trivial (and sometimes impossible) to construct an isosceles triangle with the angle $2\pi /n$. That is if you are only using ruler and compass. $\endgroup$ – Ward Beullens Jul 15 '15 at 20:43
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    $\begingroup$ It is known that it is impossible to construct an angle of $2 \pi /9$ using ruler and compass, so your approach doesn't work for the $9$-gon. $\endgroup$ – Ward Beullens Jul 15 '15 at 20:51
  • $\begingroup$ I see, that was a typo, sorry $\endgroup$ – Ward Beullens Jul 15 '15 at 20:54

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