User HyperLuminal asked for help to prove the following statement:
Connecting the feet of the altitudes of a given triangle, we obtain another triangle for with the altitudes of the original triangle are angle bisectors of the new triangle.
User @Alexey Burdin gave nice answer. Alexey’s answer is based on a fact true only in Euclidean geometry. He had a quadrangle with two opposite right angles and used the fact that this quadrangle has a special circum circle centered at the altitude point of the triangle of the OP… (Pls. see his answer.)
My claim is that the OP’s claim is true in the hyperbolic geometry as well. However I cannot prove this statement. My main problem is that I cannot detach myself from the Burdin solution.
Help needed either to falsify my claim or to prove it.
Here’s a coordinate proof using the Poincaré disk model (identified with the unit circle).
Take $A = (p,0)$, $B=(q,0)$, $C=(0,r)$ (for non-zero $p$, $q$, $r$), so that the origin is the foot of the altitude from $C$. Inverting $B$ in the unit circle naturally gives $B^\prime = (q^{-1},0)$, which allow us to find the equation of the circle (representing the hyperbolic line) through $B$ and $C$:
$$\bigcirc{B^\prime BC}:\quad x^2+y^2 – 2x \dot{q} – 2y \dot{r} + 1 = 0$$
where $\dot{q} := \frac12(q+q^{-1})$ and $\dot{r} = \frac12(r+r^{-1})$.
Inverting $A$ in the unit circle to get $A^\prime$, and inverting $A$ in $\bigcirc{B^\prime BC}$ to get $A^{\prime\prime}$ gives the equation for the circle representing the altitude from $A$:
$$\bigcirc A^\prime AA^{\prime\prime}:\quad
x^2+y^2 – 2 x \dot{p} \dot{r} + 2 y \frac{ \dot{p} \dot{q} – 1 }{\dot{r}} + 1$$
The foot of that altitude, $D$, is the intersection the two circles. Foot $E$ of the altitude from $B$ is another such intersection. Rather than solve the messy quadratics, let’s remind ourselves that we’re interested in comparing $\angle DOB$ and $\angle EOA$. Then, because hyperbolic lines through $O$ are represented by Euclidean lines, so that $D = (d, d m)$ for some slope $m$, we can substitute these coordinates into the two equations above; eliminating $d$, we get
$$m = – \frac{(p – q) (1 – p q) \dot{r}}{
2 p q \left( \dot{p} \dot{q} + \ddot{r}^2 \right)} \tag{$\star$}$$
where $\ddot{r} := \frac12(r-r^{-1})$.
Formula ($\star$) is anti-symmetric in $p$ and $q$ (equivalently, this analysis is anti-symmetric in $A$ and $B$): switching the parameters only changes the sign of the value. Consequently, the angle formed by (Euclidean or hyperbolic) lines $\overline{OD}$ and $\overline{OE}$ is bisected by $\overline{OC}$ (here, the $y$-axis). $\square$
Notes. Recall that in the Poincaré model, Euclidean distance $d$ from the origin represents hyperbolic distance $d^\star = \log\frac{1+d}{1-d}$, so that
$$d = \frac{\exp d^\star – 1}{\exp d^\star + 1} =\tanh\frac{d^\star}{2} \qquad \dot{d} = \coth d^\star \qquad \ddot{d} = -\operatorname{csch} d^\star$$
Therefore, we can ultimately write ($\star$) as
$$m = \frac{\sinh(p^\star – q^\star)\;\sinh 2r^\star}{2(\;\sinh p^\star \sinh q^\star + \cosh p^\star \cosh q^\star \sinh^2 r^\star\;)} \tag{$\star\star$}$$
where $p^\star := |OA|$, $q^\star := |OB|$, $r^\star := |OC|$ are hyperbolic lengths of elements in hyperbolic $\triangle ABC$. (We assign opposite signs to $p^\star$ and $q^\star$ if they are on opposite sides of $O$, and matching signs if they’re on the same side of $O$.) Expressing $(\star\star)$ in terms of the sides and angles of $\triangle ABC$ is left as an exercise for the reader. (As a start: $p^\star – q^\star = |AB|$. Also, as a sanity check: the formula must reduce to $\tan C$ in the “infinitesimal case”, since $\angle AOE = \angle BOD = \angle C$ in a Euclidean triangle.)
Not sure if your conjecture is true in hyperbolic geometry.
Here is a construction to test and play around with it (and inspire you for a proof or to find an counterexample)
Remember ONE counter example is enough to disprove your conjecture, but no amount of examples is enough to prove your conjecture (for proof you just need other ways)
It is a quite fiddely construction. (be warned)
Best way is to construct it in the Beltrami klein model of hyperbolic geometry (see https://en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model )
The wikipedia page gives all you the construcions you need to make a countermodel. ( I resently added them 🙂
Start with an triangle
Find the polars of the sides
Draw lines between the vertices of the triangle and the poles of the opposite sides
Construct the altitude triangle
Construct the angle bisectors (again see the wikipedia page)
and check if they do align
Good luck
EDITED
first I thought I had found a counter example, but then reconcidering maybe I made a mistake somewhere in my construction (did i not say it was fiddely ? , see comments)