This is probably an easy question, but I don't know any Riemannian geometry and a literature search hasn't helped. Any help (e.g. providing a reference) would be greatly appreciated.


For a triangle $ABC$ in euclidean space, denote by $\alpha$ the angle at $A$ and by $a$ its opposite side. Given $\epsilon >0$, consider all triangles $ABC$ with $\alpha>\epsilon$. Then it is clear (e.g. from the sine law) that the perimeter $p(ABC)$ of any such triangle satisfies $p(ABC)< C\_\epsilon \cdot a$, with $C\_\epsilon$ a positive constant only depending of $\epsilon$.


The same type of bound holds true, for the same reasons, for triangles in hyperbolic space of constant negative curvature $-\rho$, with a constant $C\_\epsilon$ depending on $\rho$. (However, note that it doesn't hold for spaces of positive curvature, as an example with two geodesics on the sphere shows).


My question: Let $X$ be a complete Riemannian manifold with sectional curvatures in $[-\rho,0]$ for some $\rho>0$. Is the analogous statement true? That is, is it true that for every $\epsilon>0$, there exists $C\_\epsilon>0$ such that $p(ABC)< C\_\epsilon \cdot a$ holds for every geodesic triangle $ABC$ with $\alpha>\epsilon$?


In fact, I am only interested in the case where $X$ is a symmetric space of non-compact type, but it seemed more reasonable to ask the question in a more general setting.


Thank you!



Bounding the perimeter of a geodesic triangle in spaces of non-positive curvature
If the manifold is complete simply connected and curvature $\le 0$ then it is [CAT(0) space](http://en.wikipedia.org/wiki/CAT%28k%29_space).
In particular, all the triangles are thin; i.e., the flat triangle with the same sides has bigger angles. 


So your question can be reduced to the question in plane geometry and the answer is yes.


