Maximum distance from top of a right angled isoscleses triangle to the nearest intouch points

Question

The post Show that any side of a hyperbolic triangle in the hyperbolic plane lies in the closed $δ$-neighborhood of the union of the two other sides. made me puzzeling about the next puzzle:

Given en Hyperbolic right angled isoscleses triangle $\triangle ABC $ where $C$ is the right angle.

What is the maximum distance of $CI_b$ where $I_b$ is the point where the inscribed circle touches side $AC$?

The maximum distance of $CI_b$ appears where $\triangle ABC $ is a Schweikart triangle https://en.wikipedia.org/wiki/Hyperbolic_triangle#Schweikart_triangle , the right angled isoscleses triangle where $A$ and $B$ are ideal points and the distances $AB$, $AC$ and $BC$ are infinite

Following https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#Distances_between_vertex_and_nearest_touchpoints the formula $CI_a= CI_b = \frac {1}{2}(BC + AC- AB)$ also seems valid in hyperbolic geometry.

But sadly I was not able to use it to calculate $CI_a$ :
$\lim {a \to \infty} (2 a - \operatorname{arcosh} ( \cosh^2 a))$ only gives chaos

Also other ideas to solve my puzzle went nowhere

But then I was able to solve my puzzle using the Poincare disk model:

Let $C =(0,0)_e$ and $A $ and $B$ the ideal points $A = (1,0)_e$ and $B = (0,1)_e$

In the Poincare disk model $AB$ is the arc of the circle centered at $(1,1)_e$ inside the unit circle.

Let $I_c$ be the point where the inscribed circle touches side $AB$

Then $ AI_c = BI_c $ , in the Poincare disk model $I_c$ is where the arc $AB$ intersects the line $ x = y$ so $I_c = \left( 1-\frac {1}{2}\sqrt{2} , 1-\frac {1}{2}\sqrt{2} \right)_e \approx (0.293 , 0.293 )_a $

Then $I_c$ , $A$ and $I_b$ are on a circle with $ y=0 $ ( hyperbolicly the horocycle centered at $A$ through $I_c$ ) in the Poincare disk model this is the circle with centre $ H_A = (2-\sqrt{2} , 0 )_e \approx (0.586 , 0)_a $

Then $d_e(I_bH_A) =d_e( H_A A)$ and $I_b$ is on $CA$

so $I_b$ is the point $ (3 - 2\sqrt{2} , 0 )_e \approx (0.172 , 0)_a $

And de distance $d_h(CI_b) = \ln ( \sqrt{2}) = \frac{1}{2} \ln ( 2) \approx 0.347 $

This solution is very reliant on the Poincare disk model and I don't really like that.

So I was wondering can I solve my puzzle without referencing to any model hyperbolic geometry ?