AAS is not valid in spherical geometry

AAS (http://planetmath.org/AAS) is not valid in spherical geometry (http://planetmath.org/SphericalGeometry). This fact can be determined as follows:

Let $\mathrm{\ell }$ be a line on a sphere and $P$ be one of the two points that is furthest from $\mathrm{\ell }$ on the sphere. (It may be beneficial to think of $\mathrm{\ell }$ as the equator and $P$ as the .) Let $A,B,C\in \mathrm{\ell }$ 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 $\mathrm{\ell }$ is also a circle having $P$ as one of its centers (http://planetmath.org/Center8) with radii $\overline{AP}$, $\overline{BP}$, and $\overline{CP}$, we have that $\overline{AP}\cong \overline{BP}\cong \overline{CP}$ and that $\mathrm{\ell }$ is perpendicular to each of these line segments. Thus, the triangles $\mathrm{△}ABP$ and $\mathrm{△}ACP$ 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, $\mathrm{△}ABP\cong ̸\mathrm{△}ACP$ because the length of $\overline{AB}$ is strictly less than the length of $\overline{AC}$.