Compare the heights of two midpoints in a hyperbolic triangle

With a slight change in notation, consider this figure:

where $b := \frac12|\overline{AC}|$ and $c := \frac12|\overline{AB}|$.

All we need is the Law of Sines and the Double-Angle Formula for Sine.

$$\begin{align} \triangle ABC\;:&\quad \frac{\sin B\;}{\sin C\;} = \frac{\sinh 2b}{\sinh 2c} = \frac{2\sinh b \cosh b}{2\sinh c\cosh c} \quad\to\quad \frac{\sin B \sinh c}{\sin C \sinh b} = \frac{\cosh b}{\cosh c} \\[4pt] \triangle BPP^\prime:&\quad \frac{\sin B\;}{\sin P^\prime} = \frac{\sinh p}{\sinh c} \quad\to\quad \sin B \sinh c = \sinh p \\[4pt] \triangle CQQ^\prime:&\quad \frac{\sin C\;}{\sin Q^\prime} = \frac{\sinh q}{\sinh b} \quad\to\quad \sin C \sinh b = \sinh q \end{align}$$ Thus,

$$\frac{\sinh p}{\sinh q} = \frac{\cosh b}{\cosh c} $$

As one might expect, the "heights" $\overline{PP^\prime}$ and $\overline{QQ^\prime}$ are congruent if and only if $\triangle ABC$ is isosceles with vertex $A$. $\square$