I am hoping that someone can help me see understand a proof that the sum of angles of a hyperbolic triangle $\triangle ABC$ is strictly less than $\pi$.
I want to stick to the upper-half-plane model $\mathbb H:=\{z\in \mathbb C : Im (z)>0\}$ of hyperbolic geometry where lines are either vertical lines or semicircles centered on (but excluding) the $x$-axis.
To simply things, by using an LFT we can take one vertex $A$ to be the point $i$ and another $B$ to be $(0,k)\in \mathbb H$ for some $k>0$.
Now, call the third vertex $C$. We have the other two sides $AC$ and $BC$ as arcs of the circles $C_1 =(x-a)^2+y^2=r_1^2$ and $C_2=(x-b)^2+y^2=r_2^2$, respectively, so that $x^2+1=r_1^2$ and $x^2+k^2=r_2^2$.
In my class notes, there is a brief mention of how this simplification can help us with the proof. However, I do not see at all how this helps us get the result that $\triangle ABC$ sums to less than $\pi$. Therefore, any help is very appreciated.
P.S., My knowledge is elementary, so I am hoping for a proof using basic tools.
Take your triangle and move it to the standard position. Now apply a Cayley transform to take the upper half plane to the unit disk. It maps circles/lines perpendicular to the real line to circles/lines perpendicular to the unit circle, and it maps $i\mapsto 0$, so your triangle will map to a triangle with one vertex at $0$.
Two of its sides will be segments that lie on diameters of the unit circle and the third will be an arc of a circle that is perpendicular to the unit circle. Then it's clear from inspection that the sum of the angles is less than $\pi$.
Because the Cayley transform is a LFT it's conformal, so the angles of the image of your triangle are the same as the images of your original triangle.