Angle Sum Property Of A Triangle & Exterior Angle Theorem

Triangle is the smallest polygon which has three sides and three interior angles. In this article, we are going to discuss the angle sum property and the exterior angle theorem of a triangle with its statement and proof in detail.

Angle Sum Property of a Triangle Theorem

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.

Angle Sum Property of a Triangle

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.

Proof for Angle Sum Property of a 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

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.

Exterior Angle Property of a 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:

Exterior Angle Property

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.

To know more about geometry visit our website BYJU’S or download BYJU’s-The Learning App from Google Play Store.

1 Comment

  1. I alerdy have tablet of byjus premium.

Leave a Comment

Your email address will not be published. Required fields are marked *