AB=AC
EC is bisector of ∠C
BD is bisector of ∠B
(i) AB=AC→∠B=∠C (isosceler triangle)
in ΔDBC and ΔECB
BC→commonside ..(1)
∠DBC=1/2∠B=1/2∠C (∵∠B=∠C)
∠ECB=1/2∠B
⇒∠DBC=∠ECB ...(2)
∠EBC=∠B=∠C=∠DBC
⇒∠EBC=∠DCB ...(3)
From (1), (2) & (3)
ΔDBC≅ΔECB
⇒BD=CE (∵ΔDBC≅ECB)
![1213143_1309896_ans_ad2b3bfbf67541a0bf2bef45798365a3.png](https://search-static.byjusweb.com/question-images/toppr_invalid/questions/1213143_1309896_ans_ad2b3bfbf67541a0bf2bef45798365a3.png)