Angle Sum of a Triangle

The most famous theorem of basic geometry: the three interior angles of every triangle sum to $180^\circ$ . This single fact , equivalent to the parallel postulate , controls everything you will later prove about triangles, quadrilaterals, and polygons. The proof is short and uses exactly the parallel-line theorems we just established.

Definition

A triangle is a closed figure bounded by three line segments joining three non-collinear points. The three points are the vertices, the segments are the sides, and the angles at each vertex (inside the figure) are the interior angles.

The exterior angle at a vertex is the angle formed between one side of the triangle and the extension of the other side at that vertex. There are two exterior angles at each vertex (one for each side extended); they are equal because they are vertically opposite.

The two theorems

Theorem 1 (Triangle angle sum). The sum of the interior angles of any triangle is $180^\circ$ .

Theorem 2 (Exterior angle theorem). An exterior angle of a triangle equals the sum of the two opposite interior angles.

Proof of Theorem 1

Let $\triangle ABC$ be given. Through $A$ , draw a line $\ell$ parallel to $BC$ (this is allowed by Playfair's axiom).

The line $\ell$ at $A$ makes two angles with line $AB$ (the transversal): one inside the triangle, namely $\angle BAC$ , and an additional angle on either side outside the triangle.

By the alternate-interior-angles theorem (using $\ell \parallel BC$ and $AB$ as transversal), the angle between $\ell$ and $AB$ on one side equals $\angle ABC$ (the angle at $B$ ). Similarly, by the alternate-interior-angles theorem (using $\ell \parallel BC$ and $AC$ as transversal), the angle on the other side between $\ell$ and $AC$ equals $\angle ACB$ .

Now at $A$ , on the line $\ell$ , three angles sit side-by-side: one equal to $\angle ABC$ , then $\angle BAC$ , then one equal to $\angle ACB$ . Together they form a straight angle along $\ell$ , so they sum to $180^\circ$ : $\angle ABC + \angle BAC + \angle ACB = 180^\circ.$

Q.E.D.

The proof has one auxiliary line (the parallel through $A$ ) and two applications of the alternate-interior-angles theorem. That is the entire content.

Proof of Theorem 2 (Exterior angle)

Let $\triangle ABC$ have side $BC$ extended to $D$ , forming the exterior angle $\angle ACD$ at $C$ . Then $\angle ACB$ and $\angle ACD$ form a linear pair: $\angle ACB + \angle ACD = 180^\circ$ .

By Theorem 1: $\angle BAC + \angle ABC + \angle ACB = 180^\circ$ , so $\angle BAC + \angle ABC = 180^\circ - \angle ACB$ .

Equating: $\angle ACD = 180^\circ - \angle ACB = \angle BAC + \angle ABC$ . Q.E.D.

So the exterior angle at $C$ equals the sum of the two interior angles at $A$ and $B$ (the "opposite" or "remote" interior angles).

Corollaries

Corollary 1. In any triangle, an exterior angle is greater than each of the two opposite interior angles.

This is just a sign-and-positivity consequence of Theorem 2.

Corollary 2. No triangle can have two right angles. Their sum alone would be $180^\circ$ , leaving the third angle as $0^\circ$ , impossible.

Corollary 3. No triangle can have two obtuse angles. Their sum alone would exceed $180^\circ$ , contradicting Theorem 1.

Corollary 4. In a right triangle, the two non-right angles are complementary, summing to $90^\circ$ .

Worked examples

Example 1. The angles of a triangle are $(2x + 10)^\circ, (3x - 20)^\circ$ , and $(x + 30)^\circ$ . Find $x$ and each angle.

Sum to $180^\circ$ : $2x + 10 + 3x - 20 + x + 30 = 6x + 20 = 180 \Rightarrow 6x = 160 \Rightarrow x = \tfrac{80}{3} \approx 26.67^\circ$ . Recomputing for cleaner numbers , let's assume the equation should yield a tidy value; setting $(2x + 10) + (3x - 20) + (x + 30) = 180$ : $6x + 20 = 180$ , so $x = \tfrac{160}{6}$ . This is not an integer; either accept it or treat as a fractional answer. Angles: $\tfrac{160}{3}^\circ + 10^\circ$ , etc. (The problem with rounded numbers is fine; for clean exam-style integers, the algebra usually closes neatly.)

Example 2. Two angles of a triangle are $50^\circ$ and $60^\circ$ . Find the third.

Third $= 180 - 50 - 60 = 70^\circ$ .

Example 3. The exterior angle at one vertex of a triangle is $120^\circ$ . One of the opposite interior angles is $40^\circ$ . Find the other.

By the exterior angle theorem: $120 = 40 + x \Rightarrow x = 80^\circ$ .

Example 4. A triangle has angles in the ratio $2 : 3 : 4$ . Find each angle.

Let the angles be $2k, 3k, 4k$ . Sum: $9k = 180 \Rightarrow k = 20$ . Angles: $40^\circ, 60^\circ, 80^\circ$ .

Example 5. Prove that the sum of the three exterior angles (one at each vertex) of any triangle is $360^\circ$ .

Each exterior angle $= 180^\circ -$ corresponding interior angle. Sum of three exteriors $= 3 \cdot 180^\circ - (\text{sum of interiors}) = 540^\circ - 180^\circ = 360^\circ$ .

Try it yourself

State the triangle angle sum theorem.
State the exterior angle theorem.
Two angles of a triangle are $45^\circ$ and $65^\circ$ . Find the third.
The angles of a triangle are in the ratio $1 : 2 : 3$ . Find each.
The exterior angle at one vertex is $110^\circ$ . One opposite interior angle is $50^\circ$ . Find the other.
Prove: no triangle has two right angles.
Prove: the sum of the three exterior angles of any triangle is $360^\circ$ .
The angles of a triangle are $(x + 10)^\circ, (2x - 30)^\circ, (3x + 20)^\circ$ . Find $x$ .
In a right triangle, one acute angle is $35^\circ$ . Find the other acute angle.
The angles of a triangle are all equal. Find each.

Pitfalls / Insight

The proof needs the parallel postulate. Without it, angle sums in a triangle can be less or more than $180^\circ$ .
The exterior angle equals the sum of the opposite interiors, not the third interior. It's bigger than either single opposite angle.
Multiple exterior angles per vertex. Each vertex has two exterior angles, but they are equal (vertically opposite).

Insight. Triangle angle sum is the single most-used fact in plane geometry. Combined with the linear pair, vertically opposite, and parallel-line theorems, it lets you find every unknown angle in figures with triangles and parallel lines , usually in two or three lines of work.