triangle equality in manifold

For a generilized triangle on a manifold, (distance can be regarded as geodesic length)it is well known that for Eucilidean Geometry，the following is true:

Consider a triangle $ABC$, $D$ is the midpoint of $BC$, then $AD\leq \frac{1}{2}(AB+AC)$

what about some other cases in manifolds. according to my knowledge, it is also true in spheres with dimension n. However I did not konw general cases.

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When you say, "distance can be regarded as geodesic length," do you mean that the distance between two points is the length of a shortest path between them? I ask because two points can be connected by a geodesic that is not shortest, and one can define triangles formed by such geodesics. –  Joseph O'Rourke Feb 22 '12 at 17:06
@Joseph O'Rourke. Oh, thanks. Sorry for my poor statement, here I means the shortest geodesic –  yaoxiao Feb 22 '12 at 23:34

This is not true for sphere; take triangle with $AB=BC=CA=\tfrac23\cdot\pi$.
Your inequality is equivalent to the condition that distance function from any point is convex. In particular, it holds in any complete simply connected manifolds with sectional curvature $\le 0$.
@Anton Petrunin Sorry for my statement for this problem, I means the shortest geodesic,your counter example is just$AB=BC=AC=\frac{\pi}{3}$ –  yaoxiao Feb 22 '12 at 23:32
@yaoxiao, I did not understand your comment. Anyway, in the example above all the points lie on a great circle, so you have $AB=BC=CA=\tfrac23\cdot\pi$, $DB=DC=\tfrac13\cdot\pi$ and $AD=\pi$. –  Anton Petrunin Feb 23 '12 at 0:12