If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.

Let ABC be an isosceles triangle with AB = AC.

Base BC is produced at points E, D respectively.

AB = AC

⇒ ∠ABC = ∠ACB = x (let) (Angles opposite to equal sides in a triangle are equal)

From figure,

⇒ ∠ACD + ∠ACB = 180° (Linear pair)

⇒ ∠ACD + x = 180°

⇒ ∠ACD = 180° - x ….(1)

From figure,

⇒ ∠ABE + ∠ABC = 180° (Linear pair)

⇒ ∠ABE + x = 180°

⇒ ∠ABE = 180° - x ….(2)

From eq.(1) and (2), we have:

⇒ ∠ABE = ∠ACD

Hence, proved that the exterior angles so formed are equal to each other.