What the Exterior Angle Theorem States
When one side of a triangle is extended, the angle formed outside the triangle is called an exterior angle. In triangle ABC, if side BC is extended to D, the exterior angle ∠ACD equals the sum of the two interior angles that are not adjacent to it:
$$\angle ACD = \angle BAC + \angle ABC.$$
In words: an exterior angle of a triangle equals the sum of its two remote interior angles. The two corner angles far from the exterior angle, ∠A and ∠B here, are the remote interior angles (also called the opposite interior angles). The angle right beside the exterior angle, ∠ACB, is the adjacent interior angle, and it is not part of the sum.
Every triangle has six exterior angles, two at each vertex, and the same rule holds at each one. The theorem appears in school as part of NCERT Class 7, Chapter 6 (The Triangle and Its Properties) and under CCSS-M 8.G.A.5. It is also one of the oldest recorded results in geometry: the equality version is contained in Proposition 1.32 of Euclid's Elements.
Why the Theorem Is True, a Proof in Two Lines
Before leaning on a result, it helps to see why it must hold, and this one follows from two facts you already know.
The first is the angle sum property: the three interior angles of any triangle add to 180°.
$$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}.$$
The second is the linear pair: the exterior angle ∠ACD and the adjacent interior angle ∠ACB sit on a straight line, so they too add to 180°.
$$\angle ACD + \angle ACB = 180^{\circ}.$$
Both right-hand sides are 180°, so the two left-hand sides are equal. Subtract the shared ∠ACB from each, and what remains is exactly the theorem:
$$\angle ACD = \angle BAC + \angle ABC.$$
There is a second, equally short route using a parallel line. Draw a line through C parallel to AB. The exterior angle splits into two pieces: one equals ∠A as alternate interior angles, the other equals ∠B as corresponding angles. The two pieces together rebuild ∠A + ∠B. Either proof lands in the same place; the linear-pair version needs the least drawing, so it is the one worth keeping.
The Exterior Angle Inequality Theorem
There is a close cousin worth meeting, because confusing the two is the most common slip on this topic. The exterior angle inequality theorem says that an exterior angle of a triangle is greater than either one of its remote interior angles, taken separately:
$$\angle ACD > \angle BAC \quad \text{and} \quad \angle ACD > \angle ABC.$$
This follows instantly from the equality: if ∠ACD = ∠A + ∠B and both ∠A and ∠B are positive, then ∠ACD has to be bigger than each of them on its own. The inequality is the part Euclid stated first (Proposition 1.16), and it holds even in geometries where the cleaner sum version does not, which is why some references treat it as the more fundamental statement. For school problems, the sum version is the workhorse; the inequality is the sanity check.
A Related Fact, the Sum of All Exterior Angles
One more property rounds out the picture. If you walk all the way around a triangle, taking one exterior angle at each vertex, the three exterior angles always add to 360°:
$$\angle_1 + \angle_2 + \angle_3 = 360^{\circ}.$$
This is true for any convex polygon, not just triangles, and it is the reason a turtle tracing the outline of a shape turns through one full circle by the time it returns to its start. It is a separate result from the theorem above, so keep them apart: one is about a single exterior angle equalling two remote ones; this one is about all the exterior angles summing to a full turn.
Examples of the Exterior Angle Theorem
With the statement, the proof, and the inequality in hand, here is the theorem doing real work. The problems move from a direct sum up to a two-step algebra setup.
Example 1 - In triangle ABC, side BC is extended to D. If ∠A = 50° and ∠B = 60°, find the exterior angle ∠ACD
By the theorem, ∠ACD = ∠A + ∠B = 50° + 60° = 90°.
Final answer: ∠ACD = 90°.
Example 2 - An exterior angle of a triangle measures 92°. One remote interior angle is 50°. A student is asked for the other remote interior angle, and also for the adjacent interior angle
A first instinct is to treat 92° as the adjacent angle and subtract from 180° straight away, writing the other remote angle as 180° − 92° = 88°. Check that against the theorem: 50° + 88° = 138°, which is nowhere near the 92° exterior angle. The two remote angles must add to the exterior angle, not to 180°, so 88° cannot be right.
The correct path uses the theorem for the remote angle and the linear pair only for the adjacent one. Remote: 92° = 50° + x gives x = 42°. Adjacent: it sits on a line with the exterior angle, so y = 180° − 92° = 88°.
Final answer: the other remote interior angle is 42°, the adjacent interior angle is 88°.
Example 3 - An exterior angle of a triangle is 120°, and one remote interior angle is 45°. Find the other remote interior angle
By the theorem, 120° = 45° + x, so x = 120° − 45° = 75°.
Final answer: the other remote interior angle is 75°.
Example 4 - In a triangle, an exterior angle is given by 20x and its two remote interior angles are (7x + 5)° and 60°. Find x
By the theorem, the exterior angle equals the sum of the remotes:
$$20x = (7x + 5) + 60 ;\Rightarrow; 20x = 7x + 65 ;\Rightarrow; 13x = 65 ;\Rightarrow; x = 5.$$
Final answer: x = 5.
Example 5 - The two remote interior angles of a triangle are equal, and the exterior angle next to the third vertex is 110°. Find each remote interior angle.
The two remotes are equal, call each x, and they sum to the exterior angle:
$$x + x = 110^{\circ} ;\Rightarrow; 2x = 110^{\circ} ;\Rightarrow; x = 55^{\circ}.$$
Final answer: each remote interior angle is 55°.
Example 6 - In triangle PQR, the exterior angle at R is 130°, and ∠P is 20° more than ∠Q. Find ∠P and ∠Q
Let ∠Q = q, so ∠P = q + 20. The two are the remote interior angles for the exterior angle at R:
$$(q + 20) + q = 130 ;\Rightarrow; 2q + 20 = 130 ;\Rightarrow; 2q = 110 ;\Rightarrow; q = 55.$$
So ∠Q = 55° and ∠P = 75°.
Final answer: ∠P = 75°, ∠Q = 55°. Check: 75° + 55° = 130°, the exterior angle, as required.
Why the Exterior Angle Theorem Matters
A school theorem earns its keep by what it saves you downstream, and this one is pure efficiency.
One-step angle chasing. In any figure full of triangles, the theorem lets you carry an angle across a triangle without solving the whole triangle. Competition and exam problems lean on this constantly, because it turns a three-step computation into a single addition.
Polygon angle sums. The companion fact, that exterior angles sum to 360°, generalises to every polygon and is how you derive that the interior angles of an n-gon sum to (n − 2) × 180° without memorising the formula.
Navigation and turning. A robot or a drone tracing a polygonal path turns through exactly one full revolution, 360°, by the time it returns to its starting heading. That total-turning idea is the exterior-angle sum in motion, and it underlies how heading is tracked in route-following code.
Triangle inequality reasoning. The inequality version was Euclid's stepping stone to proving facts about which side of a triangle is longest, since a bigger angle always faces a longer side. The exterior angle being larger than each remote interior angle is one of the hinges that argument swings on.
For a Class 7 or Class 8 student, this is the first theorem where the answer comes out faster than the long way round, and that contrast is the whole point: geometry rewards seeing the shortcut.
Where Students Trip Up on the Exterior Angle Theorem
Mistake 1: Adding the wrong pair of angles
Where it slips in: A student adds the exterior angle to the adjacent interior angle, or pairs the exterior angle with a remote one, instead of summing the two remote interiors.
Don't do this: Write exterior = remote + adjacent, or set a remote angle equal to 180° minus the exterior angle.
The correct way: The exterior angle equals the two remote interior angles added together. The adjacent interior angle is the one that makes a straight line with the exterior angle, so that pair adds to 180° instead. Label the adjacent angle on your diagram before writing any equation.
Mistake 2: Confusing the equality version with the inequality version
Where it slips in: A problem asks whether an exterior angle is bigger than a remote interior angle, and the student tries to set them equal.
Don't do this: Treat "exterior angle is greater than each remote interior angle" as "exterior angle equals a remote interior angle."
The correct way: Keep them straight. The exterior angle equals the sum of both remotes (equality theorem) and is greater than either remote on its own (inequality theorem). The second follows from the first because both remote angles are positive.
Mistake 3: Applying the triangle rule to a polygon vertex
Where it slips in: A figure shows a quadrilateral or pentagon, and the student applies "exterior equals sum of two remote interiors" at one of its vertices.
Don't do this: Assume the two-remote-angles rule holds for any polygon.
The correct way: The "equals the sum of the two remote interior angles" rule is specific to triangles. For a general polygon, what stays true is that all the exterior angles sum to 360°. Use the triangle version only on triangles.
Key Takeaways
The exterior angle theorem says an exterior angle of a triangle equals the sum of its two remote interior angles.
It follows in two lines from the angle sum property (180°) and the linear pair on the extended side.
The inequality version says the exterior angle is greater than either remote interior angle on its own.
The most common mistake is adding the adjacent angle or pairing to 180° instead of summing the two remotes.
All three exterior angles of a triangle add to 360°, a fact that generalises to every convex polygon.
Practice These Problems to Solidify Your Understanding
In triangle ABC, side AB is extended. The exterior angle is 115° and ∠C = 40°. Find the other remote interior angle.
An exterior angle of a triangle equals 4x, and its remote interior angles are (3x + 10)° and 50°. Find x.
The two remote interior angles of a triangle are in the ratio 2 : 3, and their exterior angle is 100°. Find both angles.
Answer to Question 1: 75°. Answer to Question 2: x = 60. Answer to Question 3: 40° and 60°. If Question 1 gave you anything other than 75°, check that you added the two remote angles to the exterior angle rather than subtracting from 180° (see Mistake 1).
Want a live Bhanzu trainer to walk your child through the Triangles chapter and the exterior angle theorem? Book a free demo class — online globally.
Was this article helpful?
Your feedback helps us write better content