證明：連接$BD。$
$\because$在等邊$\triangle ABC$中，$D$是$AC$的中點(diǎn)，
$\therefore \angle DBC = \frac{1}{2}\angle ABC = \frac{1}{2} \times 60^\circ = 30^\circ，$$\angle ACB = 60^\circ。$
$\because CD = CE，$
$\therefore \angle CDE = \angle E。$
$\because \angle ACB = \angle CDE + \angle E = 60^\circ，$
$\therefore \angle E = 30^\circ。$
$\therefore \angle DBC = \angle E = 30^\circ。$
$\because DM \perp BC，$即$\angle DMB = \angle DME = 90^\circ，$
在$\triangle BDM$和$\triangle EDM$中，
$\begin{cases}\angle DBC = \angle E, \\DM = DM, \\\angle DMB = \angle DME,\end{cases}$
$\therefore \triangle BDM \cong \triangle EDM (ASA)。$
$\therefore BM = EM，$即$M$是$BE$的中點(diǎn)。
