Showing that sum of angles of hyperbolic triangle is strictly less than $\pi$ 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 $\mathbb H:=\{z\in \mathbb C : Im (z)&gt;0\}$ 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 $k&gt;0$.  

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.
I think we have to start at the very basics


how do you measure/ calculate an angle between an intersecting vertical line and half circle?
How do you measure/ calculate an angle between two intersecting half circles?


That is in fact all you need to know 

But now in analytical geometry and with the simplification that the center of the half circle is always on the x-axis

Start by calculating the intersection point 

This really should be enough

Good luck