in triangle abc with side BC is produced to D if the bisector of angle abc and angle ACD meet at point E then prove that angle bec is equal to half of angle BAC

Given: ABC is a triangle,

In which 

Also, BE and CE are the angle bisector of the angles ABC and ACD,

We have to prove that: ∠BEC = 1/2 ∠BAC

Proof:

BE is the angle bisector of angle ABC,

⇒ ∠ABE = ∠EBC

And, CE is the angle bisector of angle ACD,

⇒ ∠ACE = ∠ECD

By the exterior angle theorem,

∠ACD = ∠ABC + ∠BAC

⇒ (∠ACE + ∠ECD) = (∠ABE + ∠EBC) + ∠BAC

⇒ 2∠ECD = 2∠EBC + ∠BAC

⇒ ∠BAC = 2(∠ECD - ∠EBC) ---------(1)

Now, again by exterior angle theorem,

∠ECD = ∠EBC+∠BEC

⇒ ∠BEC = ∠ECD - ∠EBC ------------(2)

By equation (1) and (2),

∠BAC = 2 ∠BEC

⇒ 1/2 ∠BAC = ∠BEC

Hence, proved.
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