Lesson Sum of the interior angles of a triangle

Sum of the interior angles of a triangle

Theorem
For any triangle in the plane the sum of its interior angles is equal to 180°.

Proof Let ABC be a given triangle in the plane (Figure 1a). Let DE be the straight line parallel to the side AB of the triangle               (Figure 1b) and passing through its vertex C. Consider angles ABD and CBE in Figure 1b. The angle ABD is congruent to the angle CAB, because these angles are alternate interior angles formed by parallel lines AC and DE and the transversal line AB (see the lesson Parallel lines under the topic Angles, complementary, supplementary angles in this site). The angle CBE is congruent to the angle ACB, because these angles are alternate interior angles formed by parallel lines AC and DE and the transversal line CB (see the lesson Parallel lines under the topic Angles, complementary, supplementary angles in this site). The sum of the angles ABD, ABC and CBE is equal to 180°: ABD + ABC + CBE = 180°, as this sum is equal to the straight angle DBE.

Figure 1a. To the Theorem on the sum     of the interior angles of a triangle

Figure 1b. To the Proof of the Theorem on the sum of the interior angles of a triangle

Replacing the angles ABD and CBE in this equality to the congruent angles CAB and ACB, we obtain the equality
CAB + ABC + ACB = 180°,
exactly what the theorem states. The proof is completed.

The proved theorem has many consequences. We consider some of them below.

For each interior angle of the triangle there are two exterior angles.           Each of these two exterior angles is formed by one side of the interior angle and by the continuation of its other side. Figure 2 shows two exterior angles of the triangle ABC at the vertex C. One of these angles is BCD. It is formed by the side BC of the triangle and by continuation CD of the triangle side AC. Another angle is ACE. It is formed by the side AC of the triangle and by continuation CE of the triangle side BC. Note that both these exterior angles are congruent, because they are vertical (see the lesson Vertical angles under the topic Angles, complementary, supplementary angles in this site).

Figure 2. Exterior angles of a triangle

Theorem
An exterior angle of a triangle is equal (is congruent) to the sum of the two angles of the triangle that are not supplemental to the given angle.

Proof
Referring to the Figure 2, the Theorem states that the exterior angle BCD is equal to the sum of angles BAC and ABC.

The proof is very simple. First, for the sum of angles BCD and BCA we have
    BCD + BCA = 180°,
as these angles are supplementary (see the lesson Angles basics under the topic Angles, complementary, supplementary angles in this site).
From the other side, the sum of the interior angles of the triangle ABC is equal to 180°:
    BAC + ABC + BCA = 180°,
as we proved it before. Comparing these two equalities, you can conclude that
    BAC + ABC = BCD,
exactly what the theorem states. The proof is completed.

Example 1
Find the third angle of the triangle if two other angles are equal to 52° and 78°.

Solution
Since the sum of three interior angles of the triangle is equal to 180°, for the third angle we have
180° - (52° + 78°) = 180° - 130° = 50°.

Example 2
Find the exterior angle of the triangle if two angles of the triangle non-adjacent to this exterior angle are equal to 47° and 82°.

Solution
Since the exterior angles of the triangle is equal to the sum of two interior angles non-adjacent to this exterior angle, this exterior angle is equal to
47° + 82° = 129°.