I have some problem in a part of proof. The question:
Let $S$ a orientable surface whit gaussian curvature $K\leq 0$. Then two geadosics $\gamma_1,\gamma_2$ that goes from $x\in S$ can't meet each other in a point $y\in S$ such that the trace of $\gamma_1$ and $\gamma_2$ form the boundary $\partial R$ os a simple region $R\subset S$.
Sketch of proof: It's proved by absurd. Suppose that the geodesics meet each other in $y$ and form the boundary of $R$. By Gauss-Bonnet, we have $$ \sum_{i=1}^{k}\int_{s_i}^{s_{i+1}} k_g(s)ds+\underset{R}{\int\int} Kd\alpha+\theta_1+\theta_2=2\pi $$ $\gamma_1$ and $\gamma_2$ are not tangent, so we have $\theta_1+\theta_2<2\pi$. And with $K\leq 0$ we conclude that $\underset{R}{\int\int} Kd\alpha+\theta_1+\theta_2<2\pi$, an absurd.
Here's my question: In the proof of Do Carmo's book, he used the fact that $\sum_{i=1}^{k}\int_{s_i}^{s_{i+1}} k_g(s)ds=0$, and i can't se this intuitively or expressing this formaly.
I have this idea: we have $\sum_{i=1}^{k}\int_{s_i}^{s_{i+1}} k_g(s)ds=\int_{\gamma_1}k_g(s)+\int_{\gamma_2}k_g(s)$, so if we consider the fact that they form a boundary of a region, we can consider $\gamma_1$ going $p\to q$ and $\gamma_2$ in the opposite way, $q\to p$, then they cancel each other. This is the unique idea until now, i'm stucked. To be honest, i don't like this argument and think is wrong.
Ps.: Sorry my english mistakes, if someone can correct i'll be grateful.