Two conjectures about infinite triangles in hyperbolic geometry

Question

I gave an answer to this question:

Hyperbolic geometry: 2/3 ideal triangle in Poincare Disk

My answer was based on the following conjectures.

Consider a doubly infinite triangle $\triangle ABC$ in hyperbolic geometry. (The figure below is given in the Klein model)

$\triangle ABC$ can be extended to triply infinite triangles two ways: $\triangle A'BC$ and $\triangle A"BC$.

Conjecture 1.

Drop perpendiculars from $A$ to $BC$ ($a$), from $C$ to $A'B$ ($c$), and from $B$ to $A"C$ ($b$). These lines are concurrent.

Conjecture 2.

The common point of $a,b,c$ is the center of the incircle of $\triangle ABC$.

I don't seem to be able to prove these conjectures. Any help would be appreciated.

Answer 1

Conjecture 1 is true by a symmetry argument. First, Triangle $ABC$ is symmetric in reflection across line $a$. That symmetry takes $A'$ to $A''$, and then it takes $b$ to $c$, and so it take $a \cap b$ to $a \cap c$. But every point on $a$ is fixed, implying $a \cap b = a \cap c$, hence $a \cap b \cap c$ is not empty.

I'm doubtful about Conjecture 2. I am particularly suspicious of the case where $A$ is close to $BC$, which implies that $A'$ is close to $B$ and $A''$ is close to $C$. If one thought the conjecture was false, one could do some calculations in that case to verify.

Answer 2

Thank you, Lee Mosher for proving the first Conjecture and for expressing doubt about the second one. The proof helped me focus on symmetries and the doubt gave me energy to do the proof of the latter one.

Let's take the following figure (illustrating the situation in the Klein model):

First, let's note that two doubly infinite triangles are congruent if their non-zero angles equal. If the non-zero angles are equal then (and only then) the altitudes are equal as well.

In order to prove that $O$ is the incenter of $\triangle ABC$ it is enough to prove that $FO$ (the altitude of $\triangle ABO$) and $OF'$ (the altitude of $\triangle A'BO$) are congruent.

As mention before, this is the case if the nonzero angles of $\triangle ABO$ and of $\triangle A'BO$ equal.

It is enough to take a look at the figure above to see that the angles in question equal.

(The key to this proof is that the perpendicular dropped from $O$ to $AA'$ halves the angle $\angle AOA'$ because $\varepsilon$ is the angle of parallelism belonging to the same segment.)