please solve the question..????

Dear Student,

In triangle ABC,
AB = AC
Therefore, $\angle ABC = \angle ACB$ ( angles opposite to equal sides are also equal ).............(1)

AB = AC
$$\begin{gathered} \Rightarrow \frac{1}{2}AB = \frac{1}{2}AC \\ \Rightarrow BD = EC (\hspace{0.17em}As D and E are mid-points )............\left(2\right) \end{gathered}$$

In triangle BDC and triangle CEB,
$\angle DBC = \angle ECB ( from 1 )$
BD = CE ( from 2)
BC = BC
So by SAS congruency rule,
$△BDC \cong △CEB$

By CPCT,
BE = CD

Regards,