Question -
Answer -
Given : In ∆ABC, AE is the bisector of vertical exterior ∠A and AE ∥ BC
To prove : ∆ABC is an isosceles
Proof: ∵ AE ∥ BC
∴ ∠1 = ∠B (Corresponding angles)
∠2 = ∠C (Alternate angle)
But ∠1 = ∠2 (∵ AE is the bisector of ∠CAD)
∴ ∠B = ∠C
∴ AB = AC (Sides opposite to equal angles)
∴ ∆ABC is an isosceles triangle