Circumference of circle is divided into n equal parts

Question

Question -

Circumference of circle is divided into n equal parts, prove that

1) points of division are vertices of regular polygon

2) tangents at points of division are sides of regular polygon

My try -

I proved first as this - because equal arcs cutoff equal chords therefore all sides will be equal...and also equal arcs subtends equal angles at centre so all interior angles will be equal ....

I am able to prove 2) for n=3,4 but not able to generalize this..

Answer 1

Rotate the figure around the center of the circle such that the first point maps to the second points. By the given equidistance, it follows that the $k$th point maps to the $(k+1)$st point (and the $n$th point to the first). Then the tangents at the $n$ points are mapped in the same cyclic fashion. i.e., the polygon formed by them has precisely the symmetry of a regular $n$-gon ...

Answer 2

Let $A$, $C$ and $E$ be tangent points. You can easily notice that $OB$ and $OD$ are angle bisectors of $AOC$ and $COE$ respectively. From your proof of 1), you know the angle $AOC$ and $COE$ are equal to $\dfrac{360^{\circ}}{n}$. Hence, this gives you the angle $BOD$ is also equal to $\dfrac{360^{\circ}}{n}$. I hope this picture and explanations clarify all the question marks in your head.

Answer 3

The figure is invariant under the rotations of the group generated by the rotation of angle $\frac{2\pi}n$, hence all sides of the polygon the circle is inscribed in are isometric.