Star Polygon Angle Sum: Another Proof

We have shown that the angle sum of a star polygon or a pentagram is 180^\circ. In this post, we discuss another proof of the star polygon angle sum theorem.

star polygon

Let ABCDE be a star polygon with angle measures a, b, c, d and e. Recall from the Remote Exterior Angle Theorem that the measure of the  exterior angle of a triangle is the sum of the measures of its two remote interior angles.

Star Polygon Exterior Angle Theorem

Remote Exterior Angle Theorem: Angle 4 = Angle 1 + Angle 2

We use this theorem to show that the angle sum of a star polygon is 180^\circ. In equation form, we want to show that

a + b + c + d + e = 180^\circ.

We start the proof by drawing ray AP. We let a = \alpha + \beta as shown in the next figure.

star polygon angle sum

By the Remote Exterior Angle Theorem,

\angle CPG = \angle \alpha + c and \angle GPD = \angle \beta + \angle d.

Therefore, by the Angle Addition Postulate,

\angle CPD = (\angle \alpha + \angle c) + (\angle \beta + \angle d) = (\angle \alpha + \angle \beta) + \angle c + \angle d.

Since \angle a = \angle \alpha + \angle \beta,

\angle CPD = \angle a + \angle c + \angle d.

Also,  \angle CPD and \angle BPE are vertical angles, so their measures are equal. Therefore,

\angle BPE = \angle a + \angle c + \angle d.

Now, if we add the interior angles of triangle BPE, its angle sum is 180^\circ. Therefore, we have

 \angle a + \angle c + \angle d + \angle b + \angle e = 180^ \circ.

 But this is the sum of the interior angles of the star polygon. Therefore, the angle sum of a star polygon is equal to 180^\circ.


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