Are all Euclidean (resp. hyperbolic) triangles bi-Lipschitz equivalent? If $T_1$ and $T_2$ are two Euclidean triangles, is there an $L$-biLipschitz map from $T_1$ to $T_2$ for some $L$? Does the same hold for (geodesic) triangles in any geometry?
 A: In Euclidean case, yes. Let $ABC$ and $DEF$ be the triangle. Translate both so that $A$ and $D$ coincide with the coordinate  origin. The vectors $AB$  and $AC$ are linearly independent, and therefore there is a linear map that transforms them to $DE$ and $DF$. This map does the job; being linear, it is obviously bi-Lipschitz. 
Important: the constant $L$ in the definition of bi-Lipschitz maps will very much depend on the shape of these triangles. E.g., if one is almost regular while the other is long and skinny, the constant $L$ will have to be large. 
For the second question: you have to be concrete about the geometry. E.g.,  a geodesic triangle on a torus can be self-overlapping. Also, on the plane with taxicab ($\ell_1$) distance, there are weird geodesic triangles with vertices $(0,0)$, $(1,0)$ and $(1,1)$. Indeed, a geodesic connecting $(0,0)$ to $(1,1)$ can be a polygonal line (with infinitely many chains) staying below the parabola $y=x^2$. Such a triangle has a cusp at $(0,0)$ and is not bi-Lipschitz equivalent to the rectilinear triangle with the same vertices.
