Question

9. The side BC of AABC is produced on both sides. Prove that the sum of
the exterior angles so formed is greater than ZA by two right. angles.​

Answer 1

First you should make a rough sketch of triangle.

∠ 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°.