## The Exterior Angle Theorem

In the angle sum of a triangle post, we have discussed that the angle sum of a triangle is $180$ degrees.  In the angle sum of a polygon post,  we also have discussed that  and that the angle sum of a polygon with $n$ sides is $180(n-2)$. For example, a pentagon has $5$ sides, so the sum of its interior angle is $180(5-2) = 180(3) = 540$ degrees.

Figure 1 – The interior and exterior angles a triangle and a quadrilateral.

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, $\alpha$ is an exterior angle. The sum of the interior angle and the exterior angle adjacent to it is always $180$  degrees (Why?).  Angles whose sum is $180$ degrees are called supplementary angles.  If two angles are supplementary, we call them a linear pair.  For example, angles $\alpha$ and $a_1$ are supplementary angles and at the same time a linear pair, so $\alpha + a_1 = 180$ degrees. Now this means, that $\alpha = 180 - a_1$. 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 $180$ as shown in Figure 1.

Observe the computation in the two diagrams.  If we let $S_t$ be the angle sum of the exterior angles of a triangle, then $S_t = (180 - a_1) + (180 - a_2) + (180 - a_3) = 540$. Rearranging the terms, we have $S_t = 540 - (a_1 + a_2 + a_3)$.  But $a_1 + a_2 + a_3$ is the sum of the interior angles of a triangle which is $180$ degrees, so $540 - (a_1 + a_2 + a_3) = 540 - 180 = 360$ 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 $S_p$ be the sum of the exterior angles of the pentagon in Figure 2. Then

$S_p =(180 - c_1)+ (180 - c_2) + (180 - c_3) +(180 - c_4) +(180 - c_5)$. Simplifying, we have $S_p = 900 - (c_1 + c_2 + c_3 + c_4 + c_5)$. But according to the angle sum theorem for polygons, $c_1 + c_2 + c_3 + c_4 + c_5 = 540$. Therefore,$900 - (c_1 + c_2 + c_3 + c_4 + c_5) = 900 - 540 = 360$ degrees.

We have three polygons – triangle, quadrilateral, pentagon – whose angle sums of exterior angles are always $360$ degrees. Now, is this true for all polygons?  Try to compute polygons up to $10$ sides and see if the sum is $360$ 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 $a_1, a_2, \cdots a_n$ be measures of the interior angles of a polygon with n sides. Let$b_1, b_2, \cdots b_n$ be measures of the exterior angles of the same polygon where all angle names with the same subscripts are adjacent angles from $a_1$ and$b_1$ all the way up through $a_n$ and $b_n$ .  We know that adjacent interior and exterior angles are supplementary angles, so this implies that their measures add up to 180 degrees. Hence,

(a1 + b1) + (a2 + b2) + … + (an + bn) = 180 + 180 + … +180 (n of them) = 180n

Regrouping the terms of the preceding equation, we have

(a1 + a2 + … + an) + (b1 + b2 + … + bn) = 180n

But the sum of the interior angles is a1 + a2 + … + an = 180(n – 2)

So,

180(n – 2) + (b1 + b2 + … + bn) = 180n

b1 + b2 + … + bn = 180n – 180(n – 2) = 360

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

Proof 2:

Let a1, a2, …, an 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 – a1, 180 – a2, …, 180 – an.

If we let S, be the sum of the measure of the exterior angles, we have

S = (180 – a1) + (180 – a2) + (180 – a3) + … + (180 – an)

= (180 + 180 + 180 + … +180 (n of them)) – a – a2 – a3– … – an

S = 180n – (a1 + a2 + a3 + … + an)

But a1 + a2 + a3 + … + an 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.

## Parallel Lines and Transversals

Given a triangle and a line segment, let us place the triangle such that one of its side and the line segment coincide. Next, we replicate the triangle and slide to the right hand side  such that two of their vertices meet as shown in Figure 1.

Figure 1 – Translated triangles have parallel corresponing sides.

The angles with the same color are basically the same angle, so we will call them corresponding angles. The sides which are parallel are also the same side, so we will call them corresponding sides.

What can you say about their corresponding angles? Their corresponding sides?

Figure 2 – The sum of the angles in point T is 180 degrees.

Next, we replicate one of the triangles, rotate it and place it between the two triangles as shown in Figure 2. Click here to rotate the triangles using GeoGebra. In the figure, we can observe and deduce the following:

1. The measure of the three adjacent angles in T is half the circle and is therefore 180 degrees. This implies that the sum of the interior angles of triangle PQT is also 180 degrees, since they are the sum of the three adjacent angles in T.
2. PQRT and QRST are parallelograms. (Explain why the opposite sides are parallel).
3. The opposite angles of parallelogram are congruent, and their opposite sides are also congruent. (Explain why).
4. We also observe that the sum of the consecutive angles of a parallelogram is 180 degrees. For example, the sum of angle PQR and angle QRT is 180 degrees.
5. The interior angles of a parallelogram consists of 2 green angles, 2 yellow angles and 2 red angles, which means that the angle sum of the interior angles of a parallelogram is twice 180 or 360 degrees.
6. PQRS is a trapezoid and the sum of its interior angles is also 360 degrees. (Explain why QR and PS are parallel).

If we extend QR and QT, then we will two parallel line segments and a transversal. In Figure 3, OR and PS are parallel lines a NU is a transversal. From Figure 1 and Figure 2, the yellow angles are corresponding angles, so RQT and PTQ are congruent.  We can also see that  the blue angle and yellow angle are supplementary (the sum of their measure is 180 degrees).

Figure 3 – NU is a transversal to the parallel lines OR and PS.

With the knowledge that the blue angle and the yellow angle are supplementary, we can deduce the following:

1. Interior angles on the same side of a transversal of parallel lines (RQT and STQ) are supplementary.
2. Angles OQN and OQT are supplementary so NQT must have the same measure as that of the yellow angle.
3. Angles RQT and RQN are supplementary so RQN must have have the same measure as that of the blue angle.
4. From 2 and 3, it follows that PTU is a blue angle and STU is a yellow angle.

With all the observations and deductions above, we came up with the following diagram. Note that angles symbols with the same color are congruent angles.

Figure 4 – Angle signs with the same color are congruent.

Pairs of these angles have special names  for distinction purposes. Most of the names are descriptive, so it is easy to determine other pairs. Some pairs of angles and their names are listed below.

1.) TQR and QTS are same side interior angles.

2.) NQR and STU are same side exterior angles.

3.) RQT and PTQ are alternate interior angles.

4.) STU and NQO are alternate exterior angles.

5.) NQO and QTP are corresponding angles.

Parallel lines and its relationship with its interval was mentioned in Euclid’s book The Elements. It is called the Fifth postulate or the Parallel Postulate. The parallel postulate states that

If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.

All the relationships we have deduced above can be derived from this single statement.

I have also written why the angle sum of the interior angles of a triangle is 180 degrees. C

## Sum of the interior angles of a polygon

We were taught that if we let $m$ be the angle sum (the total measure of the interior angles) and $n$ be the number of vertices (corners)  of a polygon, then $m = 180(n-2)$.  For example, a quadrilateral has $4$ vertices, so its angle sum is $180(4-2) = 360$ degrees.  Similarly, the angle sum of a hexagon (a polygon with $6$ sides) is $180(6-2) = 720$ degrees.

But where did this formula come from?  Does this formula work for all polygons?  (Note that in this discussion, when we say polygon, we only refer to convex polygons).

Before we answer these questions, let us first have a brief review of some elementary concepts.

Polygons and Interior Angles

A polygon is a closed figure with finite number of sides. In the figures below, $ABCDE$ is a polygon with $5$ sides and ($5$ vertices).  It is clear that the number of sides of a polygon is always equal to the number of its vertices.

A polygon has interior angles.  In the first figure below, angle $B$ measuring $91$ degrees is an interior angle of polygon $ABCDE$. The angle sum $m$ of $ABCDE$ (not drawn to scale) is given by the equation

$m = 126 + 91 + 113 + 102 + 108 = 540$ degrees.

In the second figure, if we let $a_1, a_2$ and $a_3$ be the measure of the interior angles of triangle $ABE$, then the angle sum m of triangle $ABE$ is given by the equation $m = a_1 + a_2 + a_3$.

Angle Sum

To generalize our calculation of  angle sum, we use the fact that the angle sum of a triangle is $180$ degrees. Notice that any polygon maybe divided into triangles by drawing diagonals from one vertex to all of the non-adjacent vertices.  In the second figure above, the pentagon was divided into three triangles by drawing diagonals from vertex $E$ to the non-adjacent vertices $B$ and $C$ forming $BE$ and $CE$.  Now let $a_k, b_k$ and $c_k$, where $k = 1, 2, 3$ be measures of the interior angles of the three triangles as shown on the second figure.

Calculating the angle sum of pentagon $ABCDE$ we have

Notice that the angle measures in the first line of our equation is just a rearrangement of the measures of the interior angles of the three triangles. Hence, the angle sum of the pentagon is equal to the angle sum of the three triangles. Therefore, we can conclude that the sum of the interior angles of a polygon is equal to the angle sum of the number of triangles that can be formed by dividing it using the method described above. Using this conclusion, we will now relate the number of sides of a polygon, the number of triangles that can be formed by drawing diagonals and the polygon’s angle sum.

From the table above, we observe that the number of triangles formed is $2$ less than the number of sides of the polygon.  This is true, because $n - 2$ triangles can be formed by drawing diagonals from one of the vertices to $n - 3$ non-adjacent vertices. Therefore, there the angle sum $m$ of a polygon with $n$  sides is given by the formula

$m = 180(n - 2)$

A More Formal Proof

Theorem: The sum of the interior angles of a polygon with $n$ sides is $180(n-2)$ degrees.

Proof:

Assume a polygon has $n$ sides. Choose an arbitrary vertex, say vertex $V$.  Then there are $n - 3$ non-adjacent vertices to vertex $V$.  If diagonals are drawn from vertex $V$ to all non-adjacent vertices, then $n - 2$ triangles will be formed.  The sum the interior angles of $n -2$ triangles is $180(n - 2)$. Since the angle sum of the polygon with $n$ sides is equal to the sum the interior angles of $n - 2$ triangles, the angle sum of a polygon with $n$ sides is $180(n-2)$. $\blacksquare$

Exercises:

1.)    Find the number the angle sum of a dodecagon ($12$-sided polygon).

2.)    The angle sum of a polygon is $3240$ degrees. What is the number of its sides?

3.)    The measure of one of the angles of a regular polygon is $150$. Find its number of sides.

4.)    From this, prove that the sum of the interior angles of a polygon is $360$ degrees.