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If the vertices of triangle $\Delta ABC$ are inverted about a circle $\omega$ centered at the orthocenter $H$ of $\Delta ABC$, the new triangle $\Delta A'B'C'$, is similar to the orthic triangle of $\Delta ABC$. Additionally it is rotated a half-turn about H. The diagram below shows this, however I am having trouble figuring out the proof, which is based on a separately derived equation (2.44) $$HX \cdot HA = HY \cdot HB = HZ \cdot HC.$$ (This is problem 5.3.4 of Geometry Revisited, Coxeter and Greitzer, and the equation 2.44 is on p. 37 of that book also).

Can anyone help supply the proof? I see that (2.44) is very close to saying that points A, B, C invert into X, Y, Z (but with a minus sign using directed segments), and that the minus sign accounts for the half-turn, but I don't see how to state this precisely. enter image description here

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Observe that

\begin{align} \angle{AYB} &= 90^{o} = \angle{AXB} \end{align}

So $AYXB$ is a cyclic quadrilateral. By the Power of the Point Theorem,

\begin{align} HA\times HX = HB\times HY \end{align}

Applying this logic for each pair of vertices, we get the desired.

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In the interim between when I asked this question and had no responses, I realized what I was missing so here it is (I could still use help with the half-turn part, but will "accept my own answer" as soon as I am allowed to in 2 days according to mse unless another appears).

enter image description here

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