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Why is SAS (Side-Angle-Side) Congruency holds true on a plane but not on a sphere? I am trying to understand why it is so (instead of proving/disproving the same). Please help me understand.

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    I agree with @JasonDeVito, this is a question that cannot be answered without a clear definition of triangle, and is noted in some of the answers below. And I want to emphasize that there is not an obvious choice here. Different definitions will yield different answers.2016-11-08

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If I'm not mistaken, if two triangles on a sphere are in the SAS relation, then they are congruent.

But if they're on two different spheres, of different sizes, then they're not.

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SAS congruence is true for spherical triangles. See Todhunter's book (I think it is freely available on internet)

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In the comments, Jason raised the question of the legitimacy of angles of $\pi$ radians. This may be what Jason had in mind:

Let $A,B$ be antipodal, let $C$ be any other point. Then there are infinitely many non-congruent triangles $ABC$, all sharing the sides $AC$ and $BC$ and the angle at $C$. Of course, that angle at $C$ is $\pi$.

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SAS congruence does NOT hold true on a sphere. Given any three points on a sphere, there are 8 possible triangles that can be made. Lets say there are points A, B, and C on the sphere. You can draw a line segment from A to B since they both will lie on a great circle. You can make that line the short way, or the long way, by going all the way around the circle the other way. If you do that, you've created two triangles that share SAS (Side AB, Angle ABC, and Side BC), but one triangle has a surface area of greater than half a sphere, and the other has a surface area of less than half a sphere, making them not congruent.