S.O.S. Mathematics CyberBoard

[Zeta]

Hello! I was thinking about how the sum of the angles in a triangle is always 180 and that it increases by 180 each time you add a side to the polygon.

This fact is used for a proof of the pythagorean theorem that I was looking at recently but it occurred to me that I've never seen a proof of how to calculate the sum of angles in a polygon.

No rush, but I'd be grateful if someone has one.

[Justin]

I assume you mean a regular polygon? Then the sum of the angles is , where is the number of sides.

I don't specifically know a "proof" for this, but you can try breaking up the polygon into congruent triangles...the idea is that you can always break the whole figure up into triangles that are congruent to each other...then use the fact that you know the sum of the angles of each triangle is 180.

Examples: (1) a triangle itself: 1 will do; (2) a square: 2 will do; (3) a pentagon: 3 (more difficult to see), etc. This is my thinking (unless someone has a counterexample to this train of thought).
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"Mathematicians are like lovers. Grant a mathematician the least principle, and he will draw from it a consequence which you must also grant him, and from this consequence another." Bernard Le Bovier Fontenelle (1657-1757)

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G.H. Hardy (1877-1947)

[Shadow]

So, to prove this you use the fact that the sum of the exterior angles is always 360.
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[Zeta]

[Shadow]

This is for all polygons (the exterior summing to 360) and NOT just regular polygons, so you cannot get that fact from just knowing that it's true for regular polygons (which is what you get if you assume the measure of the interior angles is known) so it's not a result of the measure of the interior angles being what-have-you.
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[alstat]

Consider triangle ABC. Construct line DCE parallel to line AB and (obviously) through C.

Now m<DCA + m<ACB + m<BCE = 180 (since D, C, E are collinear).

But m<CAB = m<DCA and m<CBA = m<BCE (alternate interior angle pairs),
and so m<CAB + m<ACB + m<CBA = 180.

[Zeta]

[skipjack]

Imagine you are walking along the triangle. Each time you reach a vertex, you need to turn through the exterior angle of that vertex. When you return to your starting point, you must have turned through a total of 360°, so that is the sum of the triangle's exterior angles. It follows that the interior angles total 180°.

The above argument is easily generalized to convex polygons. For polygons that are not convex, you would need to define "external angle" in such a way that it can be negative, but the result doesn't hold for some self-intersecting polygons.

[Denis]

Why not simply draw any rectangle: with 4 right angles, obviously contains 360 degrees;
draw a diagonal; resulting 2 triangles obviously 180 degrees each...
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[Shadow]