While playing around with Geogebra, I come up with the following statement. Is this a known property? Any reference or proof?
Let $ABC$ be a triangle with a point $D$ along the side $CB$ and a point $E$ along the side $AC$. Let $H_1$ be the orthocenter of $\triangle ABC$ and let $H_2$ be the orthocenter of $\triangle EDC$. Let $O_1$ be the intersection of the perpendicular bisectors of $AB$ and $ED$, and let $O_2$ be the intersection of the perpendicular bisectors of $BD$ and $EA$. Show that the lines $H_1 H_2$ and $O_1O_2$ are parallel.
Best Answer
$\require{begingroup} \begingroup\def\conj#1{\overline{#1}}$
Using the complex number representation of all the points, condition $H_1H_2\parallel O_1O_2$ means \begin{align} O_2-O_1&=k\,(H_1-H_2),\quad k\in\mathbb R \tag{1}\label{1} . \end{align}
Without loss of generality, using \begin{align} B&=(0,0),\quad C=(1,0),\quad A=(u,v),\quad D=(d,0) ,\\ E&=A(1-t)+Ct=(u(1-t),\, v(1-t)) ,\\ &u,v\in\mathbb{R},\quad d,t\in(0,1) \end{align}
and evaluating the intersection point $z$ of the lines through points $z_1,z_2$ and $z_3,z_4$ as
\begin{align} z(z_1,z_2,z_3,z_4)&= \frac{ (z_1-z_2)\,(\conj{z_3}\,z_4-\conj{z_4}\,z_3) - (z_3-z_4)\,(\conj{z_1}\,z_2-\conj{z_2}\,z_1) } { (z_1-z_2)\,(\conj{z_3}-\conj{z_4}) - (z_3-z_4)\,(\conj{z_1}-\conj{z_2}) } . \end{align}
We can find $k$ in \eqref{1} explicitly.
\begin{align} H_1 &= (u,\tfrac uv\,(1-u)) ,\\ H_2&= \left(u(1-t)+t,\ \frac1v\,(u-1)(d-t-u(1-t))\right) \end{align}
\begin{align} O_1&=\left( \frac{((u-1)^2+v^2)t^2+(1-v^2-(u-1)^2)t-d^2}{2(t-d)} \right. , \\ &\qquad\left. \frac{-u((u-1)^2+v^2)t^2+(u^2(u-1)+v^2(u+1))t-d(u^2-ud+v^2)} {2v(t-d)} \right) ,\\ O_2&=\left(\tfrac12\,d,\, v+\tfrac 1v\,(\tfrac12\,(d(1-u) -t((1-u)^2+v^2))-u(1-u)) \right) \end{align}
The value of $k$ is indeed real: \begin{align} k&=\frac{O_2-O_1}{H_1-H_2} = \frac{d-1+(1-t)(v^2+(1-u)^2)}{2(d-t)(1-u)} , \end{align}
hence $H_1H_2\parallel O_1O_2$.
$\endgroup$
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