I am recently studying CAT(0) spaces and I have some doubts. (this because my intuition goes against what I wrote in classroom, thus I am not sure if I am doing something wrong or I miswrote\misunderstood something)
For a three points $a,b,c \in X$, where $X$ is a CAT($0$) complete space, I can create a comparison triangle with vertixes $\bar a, \bar b, \bar c$ in $\mathbb{R}^2$. The application sending the points from the triangle in $X$ to the one in $\mathbb{R}^2$ is an isometry on each side of the triangle. Moreover, for every point $x,y$ in the triangle in $X$, I have $d(x,y)\leq d(\bar x, \bar y)$.
Now, I define the angle as
$$ \textrm{ang}_a(b,c)=\arccos((d^2(a,b)+d^2(a,c)-d^2(b,c)/(2d(a,b)d(a,c))).$$
As a direct consequence, the vertixes of the tringle in $X$ have the same width of the one in $\mathbb{R^2}$ right? Can I say some other things about the relation of these two triangles?
One more thing, I now define the Alexandroff angle as:
$$\textrm{A}_a(b,c)=\lim_{s,t \rightarrow 0} \textrm{ang}_a(\sigma(s),\tau(t) ),$$
where $\sigma$ and $\tau$ are the geodesics on the sides of the triangle. How do I prove now that $A_a(b,c)\leq \textrm{ang}_a(b,c)$ and that the same of the alexandroff angles are less than $\pi$? From my notes, it is a "direct consequence".
Thanks for your help.