What is the sum, in degrees, of the measures of the interior angles of a pentagon?
This can be proven by taking a regular n-gon (why not?), and drawing a (isosceles) triangle from the centre to one of the sides.
The centre angle will be
The inner angle of the n-gon will be just twice that, or
The sum of all inner angles will then be:
In case of a regular n-gon, each angle will be