Firstly, consider the power of $B$ with respect to $(ADME)$. This gives
$$BD \times BA = BM^2.$$
Similarly, we get
$$CE \times CA = CM^2$$
by considering the power of $C$ with respect to the same circle. As $M$ is the midpoint of $BC$, the two expressions are equal, so that
$$BD \times BA = CE \times CA. \qquad (1)$$
Next, since $B, C, E, D$ are concyclic, we have
$$AD \times AB = AE \times AC \qquad (2)$$
by considering the power of $A$. Adding (1) and (2), we get
$$AB^2 = AB \times BD + AB \times AD = AC \times CE + AC \times AE = AC^2.$$
This gives $AB = AC$.