the side BC of triangle ABC is produced on both sides prove that the sum of the exterior angles so formed is greater than angle A by two right angles
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∠ ABD + ∠ ABC = 180° ….(i) (Angle on the same line)
∠ ACE + ∠ACB = 180° ……….(ii)
Adding (i) and (ii), we get
∠ ABD + ∠ ABC + ∠ ACE + ∠ACB = 360°
∠ ABD + ∠ ACE + ( ∠ ABC + ∠ACB ) =360° …………..(iii)
Now, In ΔABC,
∠ ABC + ∠ ACB + ∠ BAC = 180°
∠ ABC + ∠ ACB = 180°- ∠ BAC
Putting the value of ∠ ABC + ∠ ACB in eq.(iii), we get
∠ ABD + ∠ ACE +180°- ∠ BAC =360°
∠ ABD + ∠ ACE = 360°- 180°+ ∠ BAC
∠ ABD + ∠ ACE = 180°+ ∠ BAC
Which proves that sum of the two exterior angles formed is greater than angle A by 180°.