If equal sides of an isosceles triangle are produced, prove that the exterior angles so formed are obtuse and equal.

In △ ABC,

⇒ AB = AC (Given that triangle is isosceles)

⇒ ∠B = ∠C = x (let) [Angles opposite to equal sides are equal]

Since, ∠B = ∠C so they cannot be equal to or greater than 90° as, sum of all the three angles in a triangle is equal to 180°.

From figure,

ABD is a straight line.

∴ ∠ABC + ∠CBD = 180°

⇒ x + ∠CBD = 180°

⇒ ∠CBD = 180° - x ……….(1)

ACE is a straight line.

∴ ∠ACB + ∠BCE = 180°

⇒ x + ∠BCE = 180°

⇒ ∠BCE = 180° - x ……….(2)

Since, x is less than 90°.

∴ 180° - x > 90°.

From equation (1) and (2),

⇒ ∠CBD = ∠BCE and are obtuse angles.

Hence, proved that if equal sides of an isosceles triangle are produced, the exterior angles so formed are obtuse and equal.