The internal bisector of ∠A of Δ BC meets BC at D and the external bisector of ∠A meets BC produced at E. Prove that 
.
Given: The internal bisector of 
A of 
ABC meets BC at D and the external bisector of 
A meets BC produced at E
Required: To prove 
Consider ΔABC, here AD is the internal angle bisector of ∠A
∴ By Angle bisector theorem we have, 
--eq(1)
Again Consider ΔABC, Now AE is the External angle bisector of ∠A
∴ By Angle bisector theorem we have, 
--eq(2)
From –eq(1) and –eq(2)
We have,
⇒ 
∴ 
Hence Proved
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