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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.

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Can you show that the reflection of the orthocenter at the line $AB$ lies on the circumcircle? In other words, if you reflect the altitudes $h_a$ and $h_b$ at the line through $AB$ you get a point $P'$ on the circumcircle. Now notice that $AHBP$ is a parallelogram and $\triangle ABP'$ and $\triangle ABP$ are congruent via a reflection at the perpendicular bisector of $AB$. – t.b. Apr 16 '12 at 1:15
up vote 1 down vote accepted

$\angle ABH = 90 - \angle CAB$, and $\angle BAH = 90 - \angle CBA$, so $\angle AHB = 180 - \angle ABH - \angle BAH = \angle CBA + \angle CAB = 180 - \angle ACB$. But $\angle APB = \angle AHB$ (since $AHBP$ is a parallelogram), so $\angle APB + \angle ACB = 180$, and therefore $ACBP$ is a cyclic quadrilateral.

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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 $C$.

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