Question

ΔABC in a isosceles triangle with AB = AC. D is a point on BC produced. ED intersects AB at E and AC at F. Prove that AF > AE.

Answer 1

∠AEF > ∠ABC   ...(Exterior angle property)
∠AEF = ∠DFC
∠ACB > ∠DFC    ...(Exterior angle property)
⇒ ∠ACB > ∠AFE
Since AB = AC
⇒ ∠ACB = ∠ABC
So, ∠ABC > ∠AFE
⇒ ∠AEF > ∠ABC > ∠AFE
that is ∠AEF > ∠AFE
⇒ AF > AE.