We assume that $AB < AC$ and that $\triangle ABC$ is acute. The other cases are solved similarly. Let $M$ and $N$ be the midpoints of $BC$ and $AC$. Let $A'$ be the reflection of point $A$ over the circumcentre $O$ of $\triangle ABC$. Let $E'$ be the foot of the $C$-altitude in $\triangle ACA'$. Because $\angle AE'C = \angle ADC = 90^{\circ}$ we have that $E'$ lies on the circle with diameter $AC$. That means that $ND = NE'$. Now we have: $$\angle MNE' = \angle MNC - \angle E'NC = \angle BAC - 2 \cdot \angle E'AC = \angle BAC - 2 \cdot (90^{\circ} - \angle AA'C) =$$$$ = \angle BAC - 180^{\circ} + 2 \cdot \angle ABC = \angle ABC - \angle ACB$$ We also have $$\angle MCE' = \angle ACE' - \angle ACM = \angle AA'C - \angle ACB = \angle ABC - \angle ACB = \angle MNE'$$so $MNCE'$ is cyclic. Now we also have $$\angle E'MN = \angle E'MC + \angle CMN = 2 \cdot \angle E'AN + \angle ABC = 2 \cdot (90^{\circ} - \angle ABC) + \angle ABC = 180^{\circ} - \angle ABC = \angle NMD$$Also, $\angle NDM = \angle NCM = \angle NE'M$, so $\triangle NDM \cong \triangle NE'M$. But then $E'$ is the reflection of $D$ over $MN = l$, so $E' \equiv E$ and $A,O,E$ are collinear.

WLOG $AB<AC$. Let $M$ and $N$ be the midpoints of $\overline{AC}$ and $\overline{BC}$ respectively, and set $O$ to be the circumcenter of $\triangle ABC$. Note that $MC=MD=ME=MA$, so $ADEC$ is cyclic with center $M$. This means that \begin{align*}\angle EAC &= \angle EDC = 90^\circ - \angle MNC \\&=90^\circ - \angle ABC = \angle DAB = \angle OAC,\end{align*}and so $A$, $E$, and $O$ are collinear. $\blacksquare$

NotABot wrote: Let $ABC$ be a triangle and $D$ be the foot of the altitude from $A$. Let $l$ be the line that passes through the midpoints of $BC$ and $AC$. $E$ is the reflection of $D$ over $l$. Prove that the circumcentre of $\triangle ABC$ lies on the line $AE$. $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Let $M$ be the midpoint of $BC$ Let $N$ be the midpoint of $AC$ Working in complex numbers, where $\odot (ABC)$ is the unit circle, let $a=a$, $b=b$ and $c=1$, then $o=0$, $m=\frac{b+1}{2}$ and $n=\frac{a+1}{2}$ $AD\perp BC:$ $$d=\frac{(b-c)\overline{a}+(\overline{b} -\overline{c})a+\overline{b} c-b\overline{c}}{2(\overline{b} - \overline{c})}$$$$\Rightarrow d=\frac{\frac{(b-1)}{a}+\frac{(1-b)a}{b}+\frac{(1-b)(b+1)}{b}}{\frac{2(1-b)}{b}}$$$$\Rightarrow d=\frac{a+b+1-\frac{b}{a}}{2}...(I)$$$E \text{ is the reflection of }D\text{ in }MN:$ $$e=\frac{(\frac{a+1}{2}-\frac{b+1}{2})\overline{d}+\overline{(\frac{a+1}{2})}(\frac{b+1}{2})-(\frac{a+1}{2})\overline{(\frac{b+1}{2})}}{(({\overline{\frac{a+1}{2}})-(\overline{\frac{b+1}{2}})})}$$$$\Rightarrow e=-ab\overline{d}+\frac{a+b+1+ab}{2}$$By $(I):$ $$\Rightarrow e=-ab(\frac{\frac{1}{a}+\frac{1}{b}+1-\frac{b}{a}}{2})+\frac{a+b+1+ab}{2}$$$$\Rightarrow e=\frac{a^2+1}{2}$$$$\Rightarrow \frac{e}{\overline{e}}=\frac{a^2+1}{\frac{1}{a^2}+1}$$$$\Rightarrow \frac{e}{\overline{e}}=a^2$$$$\Rightarrow \frac{e}{\overline{e}}=\frac{a}{\overline{a}}$$$$\Rightarrow \boxed{\textbf{A,O, and E are collinear}}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

Claim: $ADEC$ is cyclic with center $M$. Proof: Because $M$ is the midpoint of hypoteneuse $AC$ in $\triangle ADC$, it follows that $AM = MC = MD$. But from the reflection we also have $MD = ME$, so our claim is proved. $\square$. We now proceed with angle chasing. We wish to prove that $\angle EAC = \angle OAC$. Notice that $\angle OAC = 90 - \angle B$ from arcs in $(ABC)$. Now, $$\angle EAC = \frac12 \angle EMC = \frac12 \left( \angle A - \angle NME \right) = \frac12 \left( \angle A - \angle NMD \right)$$But $\angle NMD = \angle AMN - \angle AMD = (180 - \angle A) - 2 \angle C$ so it follows that $$\angle EAC = \frac12 \left( \angle A - 180 + \angle A + 2 \angle C \right) = \frac12 (180 - 2 \angle B) = 90 - \angle B = \angle OAC$$so we are done.