# The Exterior Angle Theorem

In the angle sum of a triangle post, we have discussed that the angle sum of a triangle is degrees. In the angle sum of a polygon post, we also have discussed that and that the angle sum of a polygon with* * sides is . For example, a pentagon has sides, so the sum of its interior angle is degrees.

The angle sums that we have discussed in both blogs refer to the sum of the interior angles. What about the exterior angles?

The exterior angle is formed when we extend a side of a polygon. In the triangle above, is an exterior angle. The sum of the interior angle and the exterior angle adjacent to it is always degrees (Why?). Angles whose sum is degrees are called *supplementary angles*. If two angles are supplementary, we call them a *linear pair*. For example, angles ** _{ }**and

**are supplementary angles and at the same time a linear pair, so**

_{ }**degrees. Now this means, that .**

**Therefore, if we want to compute the measure of an exterior angle adjacent to an interior angle, we can always subtract the measure of the interior angle from as shown in**

_{ }*Figure 1*.

Observe the computation in the two diagrams. If we let be the angle sum of the exterior angles of a triangle, then . Rearranging the terms, we have . But is the sum of the interior angles of a triangle which is degrees, so degrees.

Now, try calculating for the sum of the exterior angles of the quadrilateral above. What is your answer?

To verify our hunch, we will try to compute for the sum of the exterior angles of a pentagon.

Let be the sum of the exterior angles of the pentagon in* Figure 2*. Then

. Simplifying, we have . But according to the angle sum theorem for polygons, . Therefore, degrees.

We have three polygons – triangle, quadrilateral, pentagon – whose angle sums of exterior angles are always degrees. Now, is this true for all polygons? Try to compute polygons up to sides and see if the sum is degrees.

**Delving Deeper**

We know that in a polygon, the number of exterior angles is equal to the number of interior angles. Furthermore, we know that the angle sum of an interior angle and the exterior angle adjacent to each is always latex **180** degrees. If we have a polygon with **5** sides, then

** **

**interior angle sum + exterior angle sum = 180(5)**

** **

In general, this means that in a polygon with **n** sides

** **

**interior angle sum* + exterior angle sum = 180n **

** **

But the **interior angle sum = 180(n** **– 2)**.** **So, substituting in the preceding equation, we have

**180(n – 2) + exterior angle sum = 180n**

which means that the ** exterior angle sum = 180n – 180(n – 2) = 360 **degrees. More formal proofs using these arguments are shown below.

** **

**Theorem:** The sum of the measure of the exterior angles of a polygon with** n** sides is **360** degrees.

** **

**Proof 1:**

Let be measures of the interior angles of a polygon with **n **sides. Let be measures of the exterior angles of the same polygon where all angle names with the same subscripts are adjacent angles from and all the way up through and . We know that adjacent interior and exterior angles are supplementary angles, so this implies that their measures add up to 180 degrees. Hence,

**(a _{1} + b_{1}) + (a_{2} + b_{2}) + … + (a_{n} + b_{n}) = 180 + 180 + … +180 (n of them) = 180n**

Regrouping the terms of the preceding equation, we have

**(a**_{1}** + a**_{2}** + … + a**_{n}**) + (b**_{1}** + b**_{2}** + … + b**_{n}**) = 180n**

But the sum of the interior angles is** a _{1} + a_{2} + … + a_{n} = 180(n – 2)**

So, ** **

**180(n – 2) + (b _{1} + b_{2} + … + b_{n}) = 180n**

**b _{1} + b_{2} + … + b_{n} = 180n – 180(n – 2) = 360**

Therefore, the sum of the exterior angles of any polygon is equal to **360** degrees.

** **

**Proof 2:**

Let **a _{1}, a_{2}, …, a_{n}** be measures of the interior angles of the polygon with

**n**sides. Since each adjacent interior and exterior angle is a linear pair, it follows that the measure of the exterior angles adjacent to them respectively are

**180 – a**.

_{1}, 180 – a_{2}, …, 180 – a_{n}If we let **S**, be the sum of the measure of the exterior angles, we have

**S = (180 – a _{1}) + (180 – a_{2}) + (180 – a_{3}) + … + (180 – a_{n})**

**= (180 + 180 + 180 + … +180 (n of them)) – a _{ }– a_{2} – a_{3}– … – a_{n}**

** **

**S = 180n – (a _{1 }+ a_{2} + a_{3 }+ … + a_{n})**

** **

But **a _{1 }+ a_{2} + a_{3 }+ … + a_{n }**is the sum of the measures of the interior angles of a polygon with

**n**sides which equals

**180(n – 2),** so, **S = 180n – 180(n – 2) = 360, **which is want we want to show**.**

Therefore, the sum of the exterior angles of any polygon is equal to **360 **degrees.