geodesic triangle

Let $M$ be a differentiable manifold (at least two times differentiable) and $A,B,C\in M$ (not necessarily distinct). Let $x_1,x_2,x_3\in [0,\mathrm{\infty })$. Let ${\gamma }_1:[0,x_1]\to M$, ${\gamma }_2:[0,x_2]\to M$, and ${\gamma }_3:[0,x_3]\to M$ be geodesics such that all of the following hold:

• ${\gamma }_1(0)=A$;

• ${\gamma }_1(x_1)=B$;

• ${\gamma }_2(0)=A$;

• ${\gamma }_2(x_2)=C$;

• ${\gamma }_3(0)=B$;

• ${\gamma }_3(x_3)=C$.

Then the figure determined by ${\gamma }_1$, ${\gamma }_2$, and ${\gamma }_3$ is a geodesic triangle.

Note that a geodesic triangle need not be a triangle. For example, in ${ℝ}^2$, if $A=(0,0)$, $B=(1,2)$, and $C=(3,6)$, then the geodesic triangle determined by $A$, $B$, and $C$ is $\{(x,2x):x\in [0,3]\}$, which is not a triangle.

geodesic metric space (http://planetmath.org/GeodesicMetricSpace)