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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.

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Please note as Jason points out that the sum of exterior angles is $360^\circ$ for any (convex) polygon. – Calvin Lin Jan 29 '13 at 17:51

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 $360^\circ$.

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You might want to add the condition of convex, to avoid directed angles for the concave case. – Calvin Lin Jan 29 '13 at 17:52

Add $2$ more segments in the draw as to obtain triangles inside, and use that the inner angles of a triangle sums up to $180^\circ$, such as the angles filling one side of a line.

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Interior angles of a pentagon sum to $540$ degress because you can divide a pentagon into three triangles. The sum of the five interior angles and five exterior angles is obviously $5*180=900$ degrees because the sum of interior angle and exterior angle that corresponds to that interior angle is $180$ degrees. Now the sum of exterior angles is $900-540=360=4*90$ and that is what you have asked for.

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(The following is not as elementary as the other answers, but it puts the given claim in a greater perspective.)

Using the proper sign conventions the claim is true even if the pentagon is non-convex, but still without self-intersections. Hopf's Umlaufsatz (sorry, there is no official English name for this theorem) says the following:

Theorem. The total curvature of a smooth Jordan curve $\gamma$ in the plane is $\pm 2\pi$.

In the case of a polygon the curvature is concentrated in the vertices, and the total curvature is the sum of the turning angles of the tangent vector at the vertices, or $a+b+c+d+e$ in your figure.

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