A property of geodesic triangles in Alexandrov spaces

Question

Let $X$ be an $n$-dimensional Alexandrov space with curvature at least -1. Assume that at every point it has an $(n,\delta)$-strainer of length $\mu$, where $\delta$ and $\mu$ are independent of a point.

Does there exist $\sigma=\sigma(\delta, \mu)$ such that for any geodesic triangle of diameter less than $\sigma$ if one of its angles is at least $\pi-1/1000$ then the other two are less than $1/10$?

For smooth Riemannian manifolds with the injectivity radius at least $\mu$ the answer is positive as it was explained in the answer to this post: A property of geodesic triangles in manifolds with lower bounds on curvature and injectivity radius However I do not see how to generalize directly the argument.

Answer 1

First note that for any direction $\xi$ at $x$ there is a geodesic $[xz]$ of length $\tfrac{\mu}{10^n}$ which runs in the $\varepsilon$-close direction to $\xi$.

(This part follows from volume comparison. The space is locally-bi-Lipschitz-Euclidean, therefore voulme of small balls are almost as in the Euclidean space. On the other hand if there are no long $\tfrac{\mu}{10^n}$-long geodesics in the directions $\varepsilon$-close to $\xi$ then the volume of $2\cdot\tfrac{\mu}{10^n}$ is too small.)

Next note that if $\measuredangle [x^y_z]=\alpha$ and $|x-y|=|x-z|=r<\tfrac{\mu}{10^n}$ then $$|y-z|>\tfrac{r}{10}\cdot(\alpha-\varepsilon),$$ where $\varepsilon$ is a small constant depending on $n$ and $\delta$.

(It says roughly that in distance coordinates the geodesics from $x$ are nearly straight, say it can not turn by big angle.)

These two observations plus the standard comparison lead to the solution.