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.