How was the angle sum of polygons derived?

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    Angle Polygon Sum
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Discussion Overview

The discussion revolves around the derivation of the angle sum of polygons, specifically focusing on both the interior and exterior angles of non-concave polygons. Participants explore various methods and reasoning behind these geometric properties.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that the sum of the exterior angles of a non-concave polygon is 360 degrees and questions how this is derived.
  • Another participant presents a formula for the sum of the interior angles as 180n - 360, explaining the relationship between interior and exterior angles.
  • A different participant challenges the initial derivation by asking how the equation for the sum of interior angles was established, suggesting a potential circular reasoning issue.
  • One participant proposes a method involving dividing the polygon into triangles by connecting an interior point to the vertices, leading to a calculation of the interior angles based on the properties of triangles.
  • A later reply expresses appreciation for the explanation provided regarding the triangle method.

Areas of Agreement / Disagreement

Participants express differing views on the derivation of the angle sum formulas, with some questioning the foundational assumptions while others provide methods of derivation. No consensus is reached on the initial derivation of the interior angle sum.

Contextual Notes

There are unresolved questions regarding the assumptions behind the derivation of the angle sum formulas, particularly concerning the relationship between interior and exterior angles and the foundational proofs involved.

prasannapakkiam
We all know that for the angle sum of the external angles of a non-concaved polygons is 360. How is/was this derived...
 
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The sum of the interior angles is 180n - 360. Let a set Q denote n angles whose sum is 180n - 360: [ {a_{1}, a_{2} , ... , a_{n} ]. For any of these angles, the external angle is equal to 180 - a_{k}. Since there are n angles, the sum of all external angles is

S = \sum_{k = 1}^{n} 180 - a_{k}

S is obviously equal to 180n - (180n - 360) = 360
 
But... I see it as; the equation you started with was derived from the fact that the sum of exterior angles equalled 360... If not how was your starting equation derived?
 
Take any point inside the polygon. Join it to the vertices, to divide the polygon into n triangles.

The sum of the angles in a triangle is 180 (see any basic geometry textbook for a proof of that).

So the sum of the angles in all the triangles = 180n.

The sum of the angles round the interior point = 360.

So the sum of the interior angles if the polygon = 180n - 360.
 
Brilliant; thanks for that.
 

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