Prove that the external angle of a triangle is the sum of the opposite internal angles.
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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 form 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
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