geometrypolygons
How can I prove that exterior angles of a pentagon add up to four right angles
I have thought about dividing the pentagon into 3 triangles, then maybe using the exterior angle sum equal to two interior angles of the triangle. But I am unable work out a formal proof.
in $ \triangle ABC$, $\angle B=90^\circ$, $AD=DC$
For an n-gon, considering a line rotating around it rotates 360. This implies the standard result that the sum of the exterior angles in the direction of the rotation is 360 so the sum of the interior angles is n*180-360 = 180(n-2).
Therefore the sum of the exterior angles on both sides of each vertex is 720.
From your first diagram, each angle of the extended polygon is 180 minus the sum of the adjacent exterior angles, so their sum is 180*n-720 =180(n-4).
Best Answer
By dividing a convex $n$-gon into $n-2$ triangles, one sees that the interior angles sum up to $(n-2)\times 180^\circ$. Now the interior and exterior angles sum up to $n\times 180^\circ$, so the exterior angles sum up to $360^\circ$.