I will give a proof using complex numbers. Suppose the circumcenter of $ABC$ is at the origin, $0$. Let the vertices of the triangle be $A = a, B = b, C = c$. I will prove that the orthocenter, $H = h$, of this triangle is $h = a + b + c$. To verify this, we need to show that the dot product of $h-a$ and $b-c$ is zero. Note that
$$ (h-a) \cdot (b-c) = (a+b+c-a) \cdot (b-c) = (b+c) \cdot (b-c) = |b|^2 - |c|^2$$
and since we placed the circumcenter of the triangle at the origin, we have $|b| = |c|$, so $(h-a) \cdot (b-c) = 0$, as desired.
Next, the midpoint of $AB$ is $\dfrac {a+b}{2}$. Thus, the reflection of the orthocenter over the midpoint of $AB$ is the number $x$ such that $$x - \dfrac {a+b}{2} = \dfrac {a+b}{2} - (a+b+c) \Rightarrow x = a+b -(a+b+c) = -c$$ Finally, we can see that $|-c| = |c| = |a|=|b|$, so the reflection of the orthocenter over the midpoint lies on the circumcircle, as desired.
Note that we have actually proved something stronger than the original statement - not only does the reflection lie on the circumcircle, but it is also diametrically opposite to $C$.
I'm going to write an answer using vectors.
Let $O$ be the intersection point of $AC, BD$.
Let $$\vec{OA}=\vec{a}, \vec{OB}=\vec{b}, \vec{OC}=k\vec{a}, \vec{OD}=l\vec{b}$$
where $k,l\lt 0.$
Letting $E,F,G,H$ be the midpoints of $AB, BC, CD, DA$ respectively, we have
$$\vec{OE}=\frac{1}{2}\vec a+\frac 12\vec b,\vec{OF}=\frac 12\vec b+\frac k2\vec a, \vec{OG}=\frac k2\vec a+\frac l2\vec b, \vec{OH}=\frac 12\vec a+\frac l2\vec b.$$
Letting $I$ be the intersection point of $EG, FH$, there exist $m,n$ such that
$$\vec{EI}=m\vec{EG}, \vec{FI}=n\vec{FH}.$$
The former gives us
$$\vec{OI}-\vec{OE}=m\left(\vec{OG}-\vec{OE}\right)\iff \vec{OI}=(1-m)\vec{OE}+m\vec{OG}=\frac{1-m+mk}{2}\vec a+\frac{1-m+ml}{2}\vec b.$$
The latter gives us
$$\vec{OI}-\vec{OF}=n\left(\vec{OH}-\vec{OF}\right)\iff \vec{OI}=(1-n)\vec{OF}+n\vec{OH}=\frac{k-kn+n}{2}\vec a+\frac{1-n+nl}{2}\vec b.$$
Now since $\vec a$ and $\vec b$ are linearly independent, the following has to be satisfied :
$$\frac{1-m+mk}{2}=\frac{k-kn+n}{2}\ \text{and} \frac{1-m+ml}{2}=\frac{1-n+nl}{2}.$$
These give us $m=n=1/2$ since $(k,l)\not=(-1,-1).$
Hence, we get
$$\vec{OI}=\frac{k+1}{4}\vec a+\frac{l+1}{4}\vec b.$$
On the other hand, letting $P,Q$ be the midpoints of $AC, BD$, we have
$$\vec{OP}=\frac{k+1}{2}\vec a, \vec{OQ}=\frac{l+1}{2}\vec b.$$
Finally, we obtain
$$\vec{PI}=\frac 12\vec{PQ}.$$
Since this represents that $I$ is on the line $PQ$, we now know that we get what we want. Q.E.D.
P.S. If $(k,l)=(-1,-1)$, then $ABCD$ is a parallelogram, which is an easy case.
Best Answer
This solution doesn't use anything fancy.
These relationships can be observed if you drew the accurate diagram several times, which is a good first step for olympiad problems.
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