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Hence, the sum of the all three interior angles of a triangle is equal to 180°.

∠A + ∠B + ∠C = 180°
Verification: Let ABC be a triangle and side BC is extended to point D.
Thus, ∠ACD is the exterior angle between the side AC and the extended line CD. A
So, ∠ACD = ∠CAB + ∠ABC …. (i)

And ∠ACD + ∠ACB = 180° ….(ii)
(Linear pair of angles)
B C D
From (ii), ∠ACD = 180° – ∠ACB ….(iii)
Now, from (i) and (iii), we get

180° – ∠ACB = ∠CAB + ∠ABC
∠CAB + ∠ABC + ∠ACB = 180°

Or ∠A + ∠B + ∠C = 180°
Alternatively, we can verify the above result by another way.

Let us take a triangle ABC, draw a line CD parallel to the side BC and passing C A D
through the vertex A. Name the angles formed as shown in the figure. ∠4 ∠1 ∠5
Since, side BC || CD, AB and AC are transversals. Then,

∠4 = ∠2 ....(i) (Alternate interior angles)
∠2 ∠3
∠5 = ∠3 ....(ii) (Alternate interior angles) B C

Adding (i) and (ii), we get
∠4 + ∠5 = ∠2 + ∠3

On adding ∠1 both sides, we get
∠1 + ∠4 + ∠5 = ∠1 +∠2 + ∠3
Since, ∠1 + ∠4 + ∠5 = 180° (Sum of the angles on a straight line is 180°)

Therefore, ∠1 + ∠2 + ∠3 = 180°
Or ∠A + ∠B + ∠C = 180°

Let's perform one more activity to find the sum of the interior angles of a triangle.

activity
Take a piece of paper and cut out a triangle DABC (say), see in figure 1.
Make the altitude AM by folding DABC such that it passes through A, see in figure 2.
Now fold the three corners such that all the three vertices A, B and C touch at M, see in figure 3.
A A A

1. 2. 3.

A
B C
B C B M C B M C
Clearly, all the three angles form a linear pair of angles at M.
Thus, the sum of the angles of a triangle is 180°.

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