```Question 359501
How do you prove exterior angles of a polygon to equal 360? Please don't use the circle diagram, where when all the angles are pulled in to create a circle therefore it equals 360. i want the step by step way to actually prove why the angles to equal 360.

I am not sure this will be what you looking for but one way to look at this is:

A polygon with n sides has n - 2 triangles making it up, these triangles are made by drawing non-intersecting diagonals between the vertices (the corners) of the polygon.

A triangle's 3 angles add to 180, the way to prove this is to draw a parallel line to one of the sides of the triangle that has the vertex opposite that side as one of the points on that parallel line.

For example, let the triangle be ABC, the vertices labeled counter-clockwise, the parallel line runs through point A and is parallel to side BC, let angle 1 and angle 2 be the 2 angles on either side of angle A
such that angle 1 + angle A + angle 2 = 180 degrees.
Then it follows that angle 1 = angle B
and angle 2 = angle C
since they are alternate interior angles
therefore angle A + angle B + angle C = 180 degrees

polygon:
3 sides = 1 triangle, interior angles 180, extending the line segment that makes up the edge past each vertex shows that each interior angle will have a corresponding exterior angle that will be supplementary (add to it to make 180 degrees) to it, 3 vertices * 180 degrees per vertex = 540 degrees, 540 - 180 = 360, so sum of external angles is 360 degrees

4 sides = quadrilateral = 2 triangles, interior angles 360 (2 * 180), 4 * 180 = 720, 720 - 360 = 360 = sum external angles

5 sides = pentagon = 3 triangles, internal angles 540, 5 * 180 = 900.
900 - 540 = 360 = sum external angles

n sides = n-gon = n - 2 triangles,
internal angles = (n - 2) * 180 = 180n - 360
n-gon has n vertices
n * 180 = 180n
180n - (180n - 360) = 180n - 180n + 360 = 360 = sum of the external angles

therefore all polygons have 360 degrees as the sum of their external angles
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