Hyperbolic triangles getting thinner Let $\Delta$ be a right triangle in $\mathbb{H}^2$, and let us write $l$ and $d$ for the length of the two legs $L$ and $D$. Let $p$ be the point on the hypotenuse that projects to the middle point $q$ of $L$, and let us write $l'$ for the distance $d(p,q)$.
I can see intuitively that if we keep $D$ fixed and let $L$ go to infinity, then $l'$ tends to zero (contrary to the euclidean case where it stays constant), but I don't manage to prove it with some trigonometric argument.
Is there any easy way to prove it, without passing through the coordinates?
 A: You can go all the way to the extreme case. Let $V$ be the vertex where the right angle is located.
Extend the legs $L$ and $D$ off to infinity, and let their endpoints on the circle at infinity by $\lambda$ and $\delta$. You get two hyperbolic geodesic rays $\overline{V\lambda}$, $\overline{V\delta}$.
Now connect $\lambda,\delta$ by a geodesic line $\overline{\lambda\delta}$.
Altogether you get an "ideal" triangle $\triangle V \lambda \delta$.
As a point $p' \in \overline{\lambda \delta}$ approaches $\lambda$, the distance from $p'$ to $L$ goes to zero (this can be proved analytically in coordinates, but it is a very simple calculation if you use upper half-plane coordinates where the "circle at infinity" is $\mathbb R \cup \{\infty\}$, and $\lambda = \infty$, and so both $\overline{P\lambda}$ and $\overline{\lambda\delta}$ are vertical).
Your point $p$ is squeezed between $p'$ and $\lambda$, and so its distance to $\lambda$ also goes to zero.
A: A little hyperbolic right-triangle trigonometry readily shows this. With $\theta$ the angle opposite leg $L$, we have

$$\frac{\tanh\ell}{\sinh d}=\tanh\theta =\frac{\tanh\ell'}{\sinh(d/2)}
\quad\to\quad \tanh\ell' = \frac{\tanh\ell}{2\cosh(d/2)}$$
As $d$ grows (to infinity), so does $\cosh(d/2)$; thus, $\tanh\ell'$ shrinks (to zero), so that $\ell'$ does as well. $\square$
