# 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. 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. 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. 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 $180^\circ$.