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See, since AB =AC so the ΔABC is isosceles triangle.
so , ∠B = ∠C
now. D is on AC and E is on AB 
and ∠DBC= 1/2 ∠B
and, ∠ECB =1/2 ∠C
so, ∠DBC = ∠ECB, (∵ ∠B=∠C)
now, Δ BDC is congruent to ΔEBC by SAS congruency 
so, BD = CE ( cpct)
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