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  1. I have just read about the sum of interior angles of convex polygons with n sides, which is $$(n-2) × 180°$$

Then I tried to find the sum of interior angles of some concave polygons. Surprisingly, it seems that the formula above also applies.

Can anyone help me prove this?

  1. Also, can anyone tell me is there any pattern in the sum of exterior angles of concave polygons?
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  • $\begingroup$ 1. It is always possible to partition a concave polygon into a set of convex polygons. 2. Use the fact that an interior angle + exterior angle = $360°$ $\endgroup$
    – dodoturkoz
    Aug 18 '21 at 3:32
  • $\begingroup$ Have you understood the derivation of the formula (1) for convex polygons by forming (n-2) triangles? $\endgroup$
    – dodoturkoz
    Aug 18 '21 at 3:33
  • $\begingroup$ yeah. I think the sum of exterior angles of a concave polygon depends on the number of the reflex interior angles it has. Yes I understand the derivation. Thank you very much:)) best wishes!!! $\endgroup$
    – bruce mao
    Aug 18 '21 at 5:59
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Take a point at the center of the convex polygon. For each side of the polygon, form a triangle by taking two lines from the center point to the endpoints of the side. Those two lines and the side will form a triangle. Since there is one triangle per side, this will give a total of $n(180^{\circ})$. To get the sum of the internal angles we subtract the angle between the two lines emanating from the center point from each triangle. But the sum of the angles from the center point is $360^{\circ}=2(180^{\circ})$.

Therefore, the sum of the internal angles is $n(180^{\circ})-2(180^{\circ})=(n-2)180^{\circ}$

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  • $\begingroup$ "To get the sum of the internal angles we subtract the angle between the two lines emanating from the center point from each triangle." is unclear. Is this akin to a central angle in a circle? $\endgroup$
    – nickalh
    Aug 18 '21 at 4:06
  • $\begingroup$ @nickalh We partition the polygon into triangles. Each triangle consists of three angles, one of which is the angle between the two segments coming from the center point. The sum of those central angles is $360^{\circ}$. $\endgroup$
    – John Douma
    Aug 18 '21 at 4:20
  • $\begingroup$ Ohhh!! That's an incredibly beautiful and smart method!! Thank you very much! Best wishes:) $\endgroup$
    – bruce mao
    Aug 18 '21 at 5:47

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