First solution. Let $N$ be the midpoint of $CD$ and $F$ the intersection point of lines $AE$ and $MN$. Since $M$ is the midpoint of $BC$, the segment $MN$ is parallel to $BD$.

Therefore, $AE$, $BD$ are perpendicular if and only if $AF$, $MN$ are perpendicular.
Equivalently, triangles $AFN$ and $MDN$ are similar since they already share the same angle at $N$.
Since $AFN$ and $ADE$ are similar, then triangles $ADE$ and $MDN$ are similar, which is equivalent to the fact that triangles $ADM$ and $MDC$ are similar since $E$ is the midpoint of $DM$ and $N$ is the midpoint of $DC$.
In conclusion, $AE$ and $BD$ are perpendicular if and only if the median $AM$ of triangle $ABC$ is also an altitude. That is $AB = AC$.

Second solution. By coordinates or equivalently by complex numbers. Fix the origin at $C$, and the $x$-axis to be $CA$. The affixes of the points are
$$C(0), \quad A(a) \quad \text{ and } \quad B(b+ic),$$
where $a, b, c$ are positive real numbers. Hence, the affixes of the other points are
$$M\left(\frac{b}{2}+i \frac{c}{2}\right), \quad D\left(\frac{b}{2}\right) \quad \text{ and } \quad E\left(\frac{b}{2}+i \frac{c}{4}\right)$$

Therefore, the affixes of the vectors are
$$\overrightarrow{DB}\left(\frac{b}{2}+i c\right) \quad \text{ and } \quad \overrightarrow{EA}\left(a-\frac{b}{2}-i \frac{c}{4}\right) .$$
On the other hand, the lengths of the sides are equal to
$$AB=\sqrt{(a-b)^2+c^2} \quad \text{ and } \quad AC=a$$
The segments $AE$, $BD$ are perpendicular if and only if
$$\operatorname{Re}\left(\left(\frac{b}{2}+i c\right)\left(a-\frac{b}{2}+i \frac{c}{4}\right)\right)=0$$
which is equivalent to $2ab-b^2-c^2=0$, or, after a straight computation, to $AB=AC$.

Third solution. By using scalar products of vectors. Because $M$ is the midpoint of $BC$ and $E$ the midpoint of $MD$, we have
$$\overrightarrow{DM}=\frac{1}{2}(\overrightarrow{DB}+\overrightarrow{DC}), \text{ or, equivalently } \overrightarrow{BD}=2 \overrightarrow{MD}+\overrightarrow{DC}$$
and
$$\overrightarrow{AE}=\overrightarrow{AM}+\overrightarrow{ME}=\frac{1}{2}(\overrightarrow{AM}+\overrightarrow{AD}) .$$
Because $MD$ and $AC$ are perpendicular, we have
$$\overrightarrow{AD} \cdot \overrightarrow{MD}=\overrightarrow{ME} \cdot \overrightarrow{DC}=0 .$$
Therefore
$$\begin{aligned} \overrightarrow{AE} \cdot \overrightarrow{BD} & =2 \overrightarrow{AE} \cdot \overrightarrow{MD}+\overrightarrow{AE} \cdot \overrightarrow{DC} \\ & =(\overrightarrow{AM}+\overrightarrow{AD}) \cdot \overrightarrow{MD}+(\overrightarrow{AM}+\overrightarrow{ME}) \cdot \overrightarrow{DC} \\ & =\overrightarrow{AM} \cdot \overrightarrow{MD}+\overrightarrow{AM} \cdot \overrightarrow{DC}=\overrightarrow{AM} \cdot \overrightarrow{MC} \\ & =\frac{1}{2} \overrightarrow{AM} \cdot \overrightarrow{BC} \end{aligned}$$
Hence, $AE$ and $BD$ are perpendicular if and only if the median $AM$ of the triangle $ABC$ is an altitude. This is equivalent to $AB=AC$.