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, 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 \( \overleftrightarrow {PQ} \)

Figure 2 Proof of angle sum property

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

∠PAB + ∠BAC + ∠QAC = 180° ………(1)

SincePQ||BCand AB, AC are the transversals,

Therefore, ∠QAC = ∠ACB (pair of alternate angles)

Also, ∠PAB = ∠CBA(pair of alternate angles)

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

∠ACB + ∠BAC + ∠CBA= 180°

Thus, the sum of 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 sum of the two opposite interior angles of the triangle.

Figure 3 Exterior angle Property

In the given figure, 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:

Figure 4 Exterior Angle Property

In the given figure, side BCof ∆ABC is extended. A line \( \overleftrightarrow {CE} \)

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

Also,\( \overline {BA} ~||~\overline{CE}\)

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.

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