The Exterior Angle Theorem

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Learning goals

Identify an exterior angle and its two remote interior angles from a side extension
Derive the exterior angle theorem using triangle sums and linear pairs
Solve angle equations while checking the resulting triangle measures
Compare an exterior angle with each remote interior angle using positivity
Justify the total of three consistently oriented exterior turns around a triangle

Which angles belong together?

Throughout this lesson a triangle has three noncollinear vertices, so each interior angle is greater than $0^\circ$ and less than $180^\circ$ . All angles named here are non-reflex unless a turn is described explicitly.

Take triangle $ABC$ and extend $BC$ past $C$ to a distinct point $D$ . Thus $B,C,D$ lie on one line in that order. An exterior angle is the angle between a triangle side and the continuation of an adjacent side beyond their common vertex. Here it is $\angle ACD$ .

The solid blue arcs identify the two remote interior angles. The red dashed arc is the exterior angle at C; the small uncolored arc is its adjacent interior angle.

The remote interior angles for a chosen exterior angle are the triangle’s two interior angles at the other vertices. For $\angle ACD$ they are $\angle BAC$ and $\angle ABC$ . The angle $\angle ACB$ shares vertex $C$ with the exterior angle, so it is adjacent, not remote.

The rays $\overrightarrow{CB}$ and $\overrightarrow{CD}$ are opposite. With the common side $\overrightarrow{CA}$ , they make $\angle ACB$ and $\angle ACD$ a linear pair. A point placed somewhere outside a triangle does not create an exterior angle unless the required side extension is present.

The points A, B, E lie on one line in that order.

Check your understanding

Triangle ABC has AB extended beyond B to E, with A, B, E in that order. Which pair contains the remote interior angles for exterior angle CBE?

\angle BAC

and

\angle ACB

\angle ABC

and

\angle BAC

\angle CBE

and

\angle ACB

\angle ABC

and

\angle CBE

Why the two remote angles add to the exterior angle

The exterior opening equals the combined openings at the two remote corners. For instance, suppose the interior angles at $A$ and $B$ measure $37^\circ$ and $64^\circ$ . The triangle sum gives the angle at $C$ :

m\angle ACB=180^\circ-37^\circ-64^\circ=79^\circ.

The exterior angle completes a linear pair with that $79^\circ$ angle:

m\angle ACD=180^\circ-79^\circ=101^\circ.

The remote angles also total $37^\circ+64^\circ=101^\circ$ . The triangle-sum equation and the linear-pair equation both have a total of $180^\circ$ . Removing the same interior measure at $C$ leaves equal remainders.

The exterior angle theorem states that when one side of a triangle is extended beyond a vertex, the resulting non-reflex exterior angle has measure equal to the sum of the two remote interior angle measures. In the stated $B,C,D$ order,

m\angle ACD=m\angle BAC+m\angle ABC.

The exterior angle is the sum of the two remote interior angles#

First run the argument with another triangle. If the remote angles measure $42^\circ$ and $73^\circ$ , their sum is $115^\circ$ . The triangle sum makes the adjacent interior angle $65^\circ$ , and the linear pair makes the exterior angle $180^\circ-65^\circ=115^\circ$ .

Now let $a,b,c$ be the interior angle measures at $A,B,C$ , and let $e=m\angle ACD$ . The triangle sum, proved in the previous lesson, gives

$a+b+c=180^\circ.$

The side extension gives opposite rays $CB$ and $CD$ , so the linear pair gives

$c+e=180^\circ.$

Both $a+b+c$ and $c+e$ equal $180^\circ$ . Set them equal and subtract the common measure $c$ :

\begin{aligned} a+b+c&=c+e,\\ a+b&=e. \end{aligned}

The numerical choices played no part in the subtraction. The argument works for every triangle with this side extension.

Worked example 1 Recover a remote angle and its neighbor

Problem. In triangle $PQR$ , extend $QR$ beyond $R$ to $S$ , with $Q,R,S$ in that order. Given $m\angle PRS=128^\circ$ and $m\angle QPR=57^\circ$ , find $m\angle PQR$ and $m\angle PRQ$ . Give a reason for each calculation.

The exterior angle and one remote interior angle are given. The drawing records their positions, not a scale measurement.

Solution. The remote interior angles for $\angle PRS$ are at $P$ and $Q$ . The exterior angle theorem gives

\begin{aligned} 128^\circ&=57^\circ+m\angle PQR,\\ m\angle PQR&=71^\circ. \end{aligned}

At $R$ , the adjacent interior angle and the exterior angle form a linear pair:

$m\angle PRQ=180^\circ-128^\circ=52^\circ.$

The interior measures check: $57^\circ+71^\circ+52^\circ=180^\circ$ .

Check your understanding

A triangle has interior measures $a,b,c$ , and an exterior angle adjacent to $c$ has measure $e$ . The triangle sum gives $a+b+c=180^\circ$ , and the linear pair gives $c+e=180^\circ$ . Which justified move derives $e=a+b$ ?

A Replace

a+b

180^\circ

because the triangle has a straight-angle total B Set

e=c

because the two angles share a side C Equate

a+b+c

and

c+e

, then subtract

c

from both sides D Conclude

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

because both equations use that total

Turn the geometry into an equation

Identify the adjacent and remote angles before doing algebra. The remote measures add to the exterior measure. The adjacent interior and exterior pair totals $180^\circ$ ; adding the two remote measures to the exterior measure does not generally give that total.

Worked example 2 Solve and check the geometric values

Problem. Triangle $ABC$ has $BC$ extended past $C$ to $D$ , with $B,C,D$ in that order. The measures in degrees are $m\angle BAC=4x+5$ , $m\angle ABC=3x+6$ , and $m\angle ACD=9x-11$ . Find $x$ , the exterior measure, and all three interior measures.

Each expression labels its stated angle. Only the angle at C outside the triangle is the exterior angle used in the equation.

Solution. The angles at $A$ and $B$ are remote. The exterior angle theorem gives

\begin{aligned} (4x+5)+(3x+6)&=9x-11,\\ 7x+11&=9x-11,\\ 22&=2x,\\ x&=11. \end{aligned}

Combine like terms first. Subtracting $7x$ and adding $11$ to both sides then isolates $2x$ ; division by $2$ gives $x$ .

Substitution gives $m\angle BAC=49^\circ$ , $m\angle ABC=39^\circ$ , and $m\angle ACD=88^\circ$ . The adjacent interior angle is

$m\angle ACB=180^\circ-88^\circ=92^\circ.$

All three interior measures are positive and below $180^\circ$ , and $49^\circ+39^\circ+92^\circ=180^\circ$ . The exterior check is $49^\circ+39^\circ=88^\circ$ .

Check your understanding

In a triangle, two remote interior angles measure $(4x+13)^\circ$ and $(x+24)^\circ$ , and the corresponding exterior angle measures $(8x-2)^\circ$ . What is $x$ ?

7

13

39

11

Why an exterior angle exceeds either remote angle

Adding a positive angle makes a larger measure. For example, remote measures $31^\circ$ and $55^\circ$ give an exterior measure of $86^\circ$ . That is greater than $31^\circ$ and greater than $55^\circ$ .

The exterior angle inequality#

Take another triangle whose remote interior angles measure $28^\circ$ and $59^\circ$ . The exterior angle theorem gives

$28^\circ+59^\circ=87^\circ.$

Subtract each remote measure from the exterior measure:

\begin{aligned} 87^\circ-28^\circ&=59^\circ,\\ 87^\circ-59^\circ&=28^\circ. \end{aligned}

Both differences are positive, so $87^\circ>28^\circ$ and $87^\circ>59^\circ$ .

In general, use the theorem $e=a+b$ with $a>0^\circ$ and $b>0^\circ$ , since the triangle’s vertices are noncollinear. Then $e-a=b>0^\circ$ , so $e>a$ . Also $e-b=a>0^\circ$ , so $e>b$ .

This is the exterior angle inequality: a triangle’s exterior angle has greater measure than either of its remote interior angles. It compares each remote angle separately. It does not claim the exterior angle exceeds the adjacent interior angle. For the earlier remote measures $31^\circ$ and $55^\circ$ , the adjacent angle is $180^\circ-86^\circ=94^\circ$ , which is larger than the exterior angle.

Check your understanding

An exterior angle has measure $e$ , and its remote interior angles have measures $a$ and $b$ . Which explanation establishes $e>a$ for every triangle?

A The exterior angle looks wider in every sketch B The exterior angle and a always form a linear pair C

e=a+b

and

b

is positive D

e+a=180^\circ

and

a

is positive

Three corners, one full turn

At each corner of a triangular path, continue the incoming side straight ahead. The non-reflex angle from that direction to the outgoing side is an exterior turn. Traverse the three sides in order, turning the same way each time, either all left or all right.

Follow B to C to A to B. Each blue arc starts in the continued incoming direction and ends in the outgoing direction, so all three mark left turns.

For a triangle with interior measures $46^\circ$ , $63^\circ$ , and $71^\circ$ , the exterior turns are their supplements:

\begin{aligned} 180^\circ-46^\circ&=134^\circ,\\ 180^\circ-63^\circ&=117^\circ,\\ 180^\circ-71^\circ&=109^\circ. \end{aligned}

Their total is $134^\circ+117^\circ+109^\circ=360^\circ$ . The completed path also returns you to your starting direction after one full turn.

Three consistently oriented exterior turns total a full turn#

Take another triangle with interior measures $24^\circ$ , $73.5^\circ$ , and $82.5^\circ$ at $A,B,C$ , respectively. Their sum is $180^\circ$ . Each exterior turn forms a linear pair with its adjacent interior angle, so the three turns measure

\begin{aligned} 180^\circ-24^\circ&=156^\circ,\\ 180^\circ-73.5^\circ&=106.5^\circ,\\ 180^\circ-82.5^\circ&=97.5^\circ. \end{aligned}

Adding the turns gives

\begin{aligned} 156^\circ+106.5^\circ&=262.5^\circ,\\ 262.5^\circ+97.5^\circ&=360^\circ. \end{aligned}

This calculation starts with three measures of $180^\circ$ and removes the interior-angle total of $180^\circ$ :

$540^\circ-180^\circ=360^\circ.$

For arbitrary interior measures $a,b,c$ , let the exterior turns be $e_A,e_B,e_C$ . At each vertex the continuation and the interior angle form a linear pair, so $e_A=180^\circ-a$ , $e_B=180^\circ-b$ , and $e_C=180^\circ-c$ . Adding gives

\begin{aligned} e_A+e_B+e_C&=540^\circ-(a+b+c),\\ &=540^\circ-180^\circ,\\ &=360^\circ. \end{aligned}

The linear-pair relationships and the triangle sum give this total, independent of the particular interior measures. The three positive exterior measures describe the turns along the path when their directions are consistent.

You can extend the other side at a vertex and obtain a second exterior angle with the same measure. But swapping just that angle into a turning record reverses its direction. Its positive measure still supplements the interior angle; it no longer describes the chosen traversal’s turn. A list of three measures and a record of three directed turns are different pieces of information.

Worked example 3 Complete a triangular turning record

Problem. A path follows the three sides of a triangle once in order, turning right at every corner. Two exterior turns measure $112^\circ$ and $137^\circ$ . Find the third turn and the interior angle at that third corner. Explain why the sum rule applies.

Solution. The path takes one non-reflex exterior turn at each vertex, all in the same direction. Their measures therefore total $360^\circ$ :

$e=360^\circ-112^\circ-137^\circ=111^\circ.$

The third corner’s interior angle and this exterior turn form a linear pair, so the interior measure is $180^\circ-111^\circ=69^\circ$ .

The other interior measures would be $68^\circ$ and $43^\circ$ . Together with $69^\circ$ they total $180^\circ$ , confirming a valid triangle. The nearby turning figure shows the left-turn version; reversing the entire traversal gives three right turns with the same measures.

Check your understanding

A triangular path has one non-reflex exterior turn at each vertex, all to the left. Two turns measure $119^\circ$ and $126^\circ$ . What is the third turn?

65^\circ

115^\circ

245^\circ

61^\circ

Status	What supports this lesson
Accepted earlier	Incidence and angle-measure foundations, including angle addition and straight-angle measure; the parallel foundations from lesson 2.2
Proved earlier	Linear-pair sums, vertical-angle congruence, parallel-line angle relationships and their converses, and the triangle angle sum
Proved here	The exterior angle theorem, the exterior angle inequality, and the $360^\circ$ total of consistently oriented triangle exterior turns
Observed only	Angle measures estimated from a sketch without a stated value or a proved relationship

The Exterior Angle Theorem

Learning goals

Which angles belong together?

Why the two remote angles add to the exterior angle

Turn the geometry into an equation

Why an exterior angle exceeds either remote angle

Three corners, one full turn

Common mistakes

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Multiple Choice Questions (MCQ)

Core practice

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