Why are the central angles of a regular polygon equal to the exterior angles?

Question

I am currently in Geometry, doing proof, when me, my teacher, and my classmates realized that the exterior angles of a polygon are equal to the central angles of a polygon (for a regular polygon). What I want to know it why. Why is this true for any regular polygon? What proof gives this explanation? I will put a picture to help you visualize. [1. So can anyone explain why this is, and pictures to show markings would be appreciated.

Answer 1

The figure below shows a decagon.

Let us call the central angle $\angle AKB = \angle BKC = \theta$. It has measure $\dfrac{360^\circ}n$ by rotational symmetry, but this value is not needed here.

Since both $\triangle AKB$, $\triangle BKC$ are isosceles, we have

$$\angle KAB = \angle KBA = \angle KBC = \angle KCB = \frac12(180^\circ - \theta)$$

So we have, by adjacent angles on straight lines:

$$\angle LBC = 180^\circ - \angle KBA - \angle KBC = 180^\circ - (180^\circ - \theta) = \theta$$

proving our assertion.

If a rigorous proof is not needed, try to rotate the regular polygon. When $AB$ is rotated with respect to the center $K$ to $BC$, the angle of rotation is the center angle $\theta$. At the same time, the angle rotated is also the angle formed by $AB$ and $BC$, which is the exterior angle $\angle LBC$. (This argument can also be extended to a proof of the sum of exterior angles of a convex polygon theorem)

Answer 2

Hint. Notice that the sum of the interior angles of an $n$-gon is $180^\circ\cdot (n-2)$. Triangulate the $n$-gon and use induction in order to prove it.

Then, the sum of the exterior angles is $$180^\circ\cdot n-180^\circ\cdot (n-2)=360^\circ$$

Notice that this works for non-regular polygons too.

If you want to prove it for regular polygons without using this well-known formula, you may also approach it this way: consider you regular decagon (even though this is valid for any regular $n$-gon)

Due to symmetry, both of the smaller green angles are not only equal, but also add up to an "entire" green angle, which you see at the bottom. Since the sum of angles in a triangle is $180^\circ$, which is the same as one flat angle, we find that the two blue angles are the same. Can you take it from here?