My question is about the first two pages of Chapter II.3 of Bridson/Haefliger. Below $M^2_{\kappa}$ refers to the model 2-dimensional plane of curvature $\kappa$ (e.g., $S^2$ for $\kappa=1$ or $H^2$ for $\kappa=-1$). My difficulties disappear when $\kappa \le 0$, so you can ignore this case.
In the proof of Prop. 3.1 they say that for a CAT($\kappa$) space "one can take comparison triangles in $M^2_{\kappa}$ instead of $E^2$ in the definition of the Alexandrov angle", and refer to I.2.9 for this. However, the latter only affirms that the angle of a triangle in $M^2_{\kappa}$ is equal to its Alexandrov angle. I don't understand how this implies their claim.
Similarly, in Prop. 3.5, they consider a geodesic segment $c \colon [0,\epsilon] \to X$ and a point $y\neq p$ in a CAT($\kappa$) space $X$, and assert that $$ \lim_{s \to 0} \bar \angle_p(c(s),y)=\angle_p(c(\epsilon),y), $$ where the former refers to a limit of the corresponding comparison angles in $E^2$ and the latter to the Alexandrov angle. In the proof they observe that $s \mapsto \angle_p^{(\kappa)}(c(s),y)$ is non-decreasing (which follows immediately from the CAT($\kappa$) inequality) and define $\gamma = \lim_{s \to 0} \angle_p^{(\kappa)}(c(s),y)$. My difficulties start when they assert that the latter is equal to $\lim_{s \to 0} \bar\angle_p(c(s),y)$ referring again to I.2.9; I don't understand why I.2.9 allows one to conclude this. Then they continue to assert that the latter limit is, "by definition, the strong upper angle between $[p,y]$ and $c$". However, this is not the definition at all, which in this situation would be $$ \lim_{\delta \to 0} \sup_{s \in [0,\delta], t \in [0,a]} \bar\angle_p(c(s),c'(t)), $$ where $c' \colon [0,a] \to X$ is the geodesic joining $p$ to $y$. Why do the two coincide in this case?
I have tried to attach copies of the relevant pages of the book, but I am not sure if this will work (this is my first post).
The proposition given in the book does not help too much. There is a better proof given in the Book by Anton Petrunin:
First note that the area of a hyperbolic triangle, is proportional to its defect and the area of a spherical triangle is proportional to its excess. This gives the formula $\text{area}\Delta = (\pi -\alpha - \beta - \gamma)\frac{1}{-\kappa}$ for a triangle in some space with constant sectional curvature $\kappa$.
We therefore have:
$$\begin{align*} -\kappa \cdot \text{area}_\kappa \Delta + \kappa' \cdot \text{area}_{\kappa'} \Delta' &= (\pi - \alpha - \beta - \gamma) - (\pi - \alpha' - \beta' - \gamma')\\ &= \alpha' - \alpha + \beta' - \beta + \gamma' -\gamma \end{align*}$$
We can use this to approximate the difference of definitions of Alexandrov angle for small distances:
$$|\limsup_{t,t' \to 0}\angle^{(\kappa)}_p(c(t),c'(t')) -\limsup_{t,t' \to 0}\angle^{(\kappa')}_p(c(t),c'(t'))| = 0$$ because (for $\kappa \geq \kappa'$)
$$\begin{align*} |\angle^{(\kappa)}_p(c(t),c'(t')) - \angle^{(\kappa')}_p(c(t),c'(t'))| &\leq |\angle^{(\kappa)}_p(c(t),c'(t')) + \angle^{(\kappa)}_{c(t)}(p,c'(t')) +\angle^{(\kappa)}_{c'(t')}(c(t),p) \\ &- \angle^{(\kappa')}_p(c(t),c'(t')) - \angle^{(\kappa')}_{c(t)}(p,c'(t')) -\angle^{(\kappa')}_{c'(t')}(c(t),p)| \\ &= |\text{area}^{(\kappa)}\Delta(p,c(t),c'(t')) - \text{area}^{(\kappa')}\Delta(p,c(t),c'(t'))| \end{align*} $$ But this clearly goes to $0$ as $t,t' \to 0$.