In triangle ABC we have,

A + B + C = 180

Let B = x and C = y then,

A + 2x + 2y = 180 (BE and CE are the bisector of angles B and C respectively.)

x + y + A = 180

A = 180 – (x + y) ………….(i)

Now, in triangle BEC we have,

B = x/2

C = y + ((180 – y) / 2)

= (180 + y) / 2

B + C + BEC = 180

x/2 + (180 + y) / 2 + BEC = 180

BEC = (180 – x – y) /2 ………..(ii)

From eq (i) and (ii) we get,

BEC = A/2