Angle Sum Property of a Triangle & Exterior Angle Theorem

Angle Sum Property of a Triangle

Triangle is the smallest polygon which has three sides and three interior angles.

In the given triangle, ∆ABC, AB, BC, and CA represent three sides. A, B and C are the three vertices and ∠ABC, ∠BCA and ∠CAB are three interior angles of ∆ABC.

Figure 1 Triangle ABC

Theorem 1: Angle sum property of triangle states that the sum of interior angles of a triangle is 180°.

Proof: Consider a ∆ABC, as shown in the figure below. To prove the above property of triangles, draw a line \( \overleftrightarrow {PQ} \) parallel to the side BC of the given triangle.

Since PQ is a straight line, it can be concluded that:

∠PAB + ∠BAC + ∠QAC = 180°  ………(1)

SincePQ||BC and AB, AC are transversals,

Therefore, ∠QAC = ∠ACB (a pair of alternate angle)

Also, ∠PAB = ∠CBA (a pair of alternate angle)

Substituting the value of ∠QAC and∠PAB in equation (1),

∠ACB + ∠BAC + ∠CBA= 180°

Thus, the sum of the interior angles of a triangle is 180°.

Exterior Angle Property of a Triangle:

Theorem 2: If any side of a triangle is extended, then the exterior angle so formed is the sum of the two opposite interior angles of the triangle.

In the given figure, the side BC of ∆ABC is extended. The exterior angle ∠ACD so formed is the sum of measures of ∠ABC and ∠CAB.

Proof: From figure 3, ∠ACB and ∠ACD forms a linear pair since they represent the adjacent angles on a straight line.

Thus, ∠ACB + ∠ACD = 180°  ……….(2)

Also, from the angle sum property it follows that:

∠ACB + ∠BAC + ∠CBA = 180° ……….(3)

From equation (2) and (3) it follows that:

∠ACD = ∠BAC + ∠CBA

This property can also be proved using concept of parallel lines as follows:

In the given figure, side BCof ∆ABC is extended. A line \( \overleftrightarrow {CE} \) parallel to the side AB is drawn, then: Since \( \overline {BA} ~||~\overline{CE}\) and \( \overline{AC}\) is the transversal,

∠CAB = ∠ACE   ………(4) (Pair of alternate angles)

Also, \( \overline {BA} ~||~\overline{CE}\) and \( \overline{BD}\)iis the transversal

Therefore, ∠ABC = ∠ECD  ……….(5) (Corresponding angles)

We have, ∠ACB + ∠BAC + ∠CBA = 180° ………(6)

Since,the sum of angles on a straight line is 180°

Therefore, ∠ACB + ∠ACE + ∠ECD = 180° ………(7)

Since, ∠ACE + ∠ECD = ∠ACD(From figure 4)

Substituting this value in equation (7);

∠ACB + ∠ACD = 180° ………(8)

From the equations (6) and (8) it follows that,

∠ACD = ∠BAC + ∠CBA

Hence it can be seen that the exterior angle of a triangle is equals to the sum of its opposite interior angles.