SOLUTION: can you write me a two column proof for the polygon exterior angle-sum theorem? Im using a pentagon. thank-you!

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Question 160135: can you write me a two column proof for the polygon exterior angle-sum theorem? Im using a pentagon. thank-you!
Answer by gonzo(654)   (Show Source): You can put this solution on YOUR website!
not sure if this answers your question, but here is a proof.
if it's not what you want, it might give you some ideas of how to get what you want.
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let IA = internal angle of a polygon.
let EA = external angle of that polygon
polygon exterior angle sum theorem states that the sum of the exterior angles of any polygon is 360 degrees.
polygon interior angle sum theorem states that the sum of the interior angles of a polygon is given by the equation (n-2)*180.
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the assumption in the following proof is that the polygon interior angle sum theorem is given.
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start of proof
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statement:
Sum(IA) = (n-2)*180
proof:
this is given.
---
statement:
IA = ((n-2)*180)/n
proof:
each internal angle of a regular polygon is the sum of the internal angles divided by the number of angles.
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statement:
EA = 180 - IA
proof:
external angle of a polygon is a supplement of the corresponding internal angle.
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statement:
EA = 180 - ((n-2)*180)/n
proof:
substitution of IA with formula for IA since they are equal to each other.
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statement:
n*EA = 360
proof:
formula for EA is 180 - ((n-2)*180)/n
multiply both sides of equation by n to remove denominator.
n*EA = n*(180 - ((n-2)*180)/n)
n*EA = n*180 - (n-2)*180
n*EA = n*180 - n*180 + 2*180
n*EA = 2*180
n*EA = 360
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statement:
sum(EA) = 360
proof:
n*(EA) is the same as the sum of EA for any regular polygon with n sides by definition.
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end of proof
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using pentagon as an example, we get
n = 5
n-2 = 3
sum(IA) = 3*180 = 540
IA = 540/5 = 108
EA = 180 - 108 = 72
sum(EA) = 5*(EA) = 5*72 = 360
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using decagon (10 sides) as an example, we get
n = 10
n - 2 = 8
sum(IA) = 8*180 = 1440
IA = 1440 / 10 = 144
EA = 180 - 144 = 36
sum(EA) = 10*(EA) = 360
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