How to prove that a point equidistant from each vertex of a regular polygon (the circumcentre) exists? How do you prove that a point equidistant from each vertex of a regular polygon exists? I am told to prove this using congruent triangles.
 A: I suggest a proof by construction: basically suggest a scheme for creating a point and then show that it must be equidistant from each of the vertices.
I would start by picking two adjacent vertices, let's say $A$ and $B$, drawing angular bisectors from both, and defining the intersection point as $O$. Because all of the interior angles in a regular polygon are equal, when we bisect them the halves of the angles must be equal, so $\angle OAB = \angle OBA$. This means our triangle is isosceles, so we have that $\overline{OA}=\overline{OB}$.
Now, let's define $C$ as the other adjacent vertex to $B$. Because $\overline{OB}$ is defined as an angular bisector to $\angle ABC$, we have $\angle OBC = \angle OBA$, and because the polygon is regular we have $\overline{BC} = \overline{AB}$. Using these facts you should be able to prove that triangles $OAB$ and $OBC$ are congruent by SAS, which means that $\overline{OC} = \overline{OB}$, and you should be set up to repeat the same procedure with the next vertex.
Hope this helps!
