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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.
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Solution
∠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.
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