Preselection tests for the full-time training

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 $$C(0),A(a)\text{ and }B(b+ic),$$ where $a,b,c$ are positive real numbers. Hence, the affixes of the other points are $$M\left(\frac{b}{2}+i\frac{c}{2}\right),D\left(\frac{b}{2}\right)\text{ and }E\left(\frac{b}{2}+i\frac{c}{4}\right)$$

Therefore, the affixes of the vectors are $$\overrightarrow{DB}\left(\frac{b}{2}+ic\right)\text{ and }\overrightarrow{EA}\left(a-\frac{b}{2}-i\frac{c}{4}\right).$$ On the other hand, the lengths of the sides are equal to $$AB=\sqrt{(a-b{)}^2+c^2}\text{ and }AC=a$$ The segments $AE$, $BD$ are perpendicular if and only if $$\mathrm{Re}\left(\left(\frac{b}{2}+ic\right)\left(a-\frac{b}{2}+i\frac{c}{4}\right)\right)=0$$ 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 $$\begin{array}{rl}\overrightarrow{AE}\cdot \overrightarrow{BD}& =2\overrightarrow{AE}\cdot \overrightarrow{MD}+\overrightarrow{AE}\cdot \overrightarrow{DC}\\ & =(\overrightarrow{AM}+\overrightarrow{AD})\cdot \overrightarrow{MD}+(\overrightarrow{AM}+\overrightarrow{ME})\cdot \overrightarrow{DC}\\ & =\overrightarrow{AM}\cdot \overrightarrow{MD}+\overrightarrow{AM}\cdot \overrightarrow{DC}=\overrightarrow{AM}\cdot \overrightarrow{MC}\\ & =\frac{1}{2}\overrightarrow{AM}\cdot \overrightarrow{BC}\end{array}$$ Hence, $AE$ and $BD$ are perpendicular if and only if the median $AM$ of the triangle $ABC$ is an altitude. This is equivalent to $AB=AC$.