Prove that the sum of the measures of the interior angles of a triangle is 180°

Solution:

We have to prove that the sum of measures of the interior angles of a triangle is 180°.

Consider the triangle ABC.

Draw a line PQ parallel to BC.

From the figure,

∠BAC = ∠A

∠ABC = ∠B

∠ACB = ∠C

We know that the alternate interior angles are of equal magnitude.

∠PAB = ∠B

∠QAC = ∠C(alternate interior angles)

Now, ∠PAB + ∠BAC + ∠QAC = 180°

Where, PQ line is parallel to side Bc touching vertex A.

∠PAB = ∠ABC

∠CAQ = ∠ACB (because alternate interior angles are congruent

So, ∠ABC + ∠BAC + ∠ACB = 180°

∠B + ∠A + ∠C = 180°

Therefore, ∠A +  ∠B + ∠C = 180° is proved using the concept of angles on straight lines.

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Prove that the sum of the measures of the interior angles of a triangle is 180°

Summary:

The sum of the measures of the interior angles of a triangle is 180°.