User HyperLuminal asked for help to prove the following statement:
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.)