For triangle $AB$C with orthocenter $H$, let midpoint of $AB$ be $C'$. Let point $P$ be such that $PH$ has $PC' = C'H$. That is, $P$ is reflection of orthocenter about midpoint of $AB$.
Show $P$ lies on circumcircle of $ABC$.
I approached this by trying to show that $\angle ABC = \angle APC$. I showed that $\angle ABC$ is same as angle between lines $HA$ and $CH$ but I can't get any further. I don't know how to use the fact that orthocenter is reflecting across the midpoint.
Best Answer
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 $C$.