AAS is not valid in spherical geometry. This fact can be determined as follows:

Let $\ell$ be a line on a sphere and $P$ be one of the two points that is furthest from $\ell$ on the sphere. (It may be beneficial to think of $\ell$ as the equator and $P$ as the north pole.) Let $A,B,C \in \ell$ such that

• $A$ , $B$ , and $C$ are distinct;
• the length of $\overline{AB}$ is strictly less than the length of $\overline{AC}$ ;
• $A$ , $B$ , and $P$ are not collinear;
• $A$ , $C$ , and $P$ are not collinear;
• $B$ , $C$ , and $P$ are not collinear.

Connect $P$ to each of the three points $A$ , $B$ , and $C$ with line segments. (It may be beneficial to think of these line segments as longitudes.)

Since $\ell$ is also a circle having $P$ as one of its centers with radii $\overline{AP}$ , $\overline{BP}$ , and $\overline{CP}$ , we have that $\overline{AP} \cong \overline{BP} \cong \overline{CP}$ and that $\ell$ is perpendicular to each of these line segments. Thus, the triangles $\triangle ABP$ and $\triangle ACP$ have two pairs of angles congruent and a pair of sides congruent that is not between the congruent angles (actually, two pairs of sides congruent, neither of which is in between the congruent angles). On the other hand, $\triangle ABP \not\cong \triangle ACP$ because the length of $\overline{AB}$ is strictly less than the length of $\overline{AC}$ .