Question

side BC of a triangle ABC is produced in the both the directions. Prove that the sum of the two exterior angles so formed is greater then 180 degree

Answer 1

Step-by-step explanation:

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°.