PlanetMath: AAS is not valid in spherical geometry

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AAS is not valid in spherical geometry

(Result)

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

Let $\ell$ be a line on a sphere and 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 as the north pole.) Let $A,B,C \in \ell$ such that

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

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

$\begin{pspicture}(-1,-2)(6,5) \psarc(2.5,5){5.59017}{243.435}{296.565} \psarc(-0... ...{$B$} \rput[l](5.2,0){$C$} \psdots(0,0)(3.5,-0.5)(5,0)(2.5,4.33) \end{pspicture}$

Since $\ell$ is also a circle having 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 $\overline{AC}$ .