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AAS is not valid in spherical geometry∗
Wkbj79†
2013-03-21 23:08:38
AAS is not valid in spherical geometry. This fact can be determined as
follows:
Let ` be a line on a sphere and P be one of the two points that is furthest
from ` on the sphere. (It may be beneficial to think of ` as the equator and P
as the north pole.) Let A, B, C ∈ ` such that
• A, B, and C are distinct;
• the length of AB is strictly less than the length of 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.)
∗ hAASIsNotValidInSphericalGeometryi created: h2013-03-21i by: hWkbj79i version:
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1
P
A
C
`
B
Since ` is also a circle having P as one of its centers with radii AP , BP ,
and CP , we have that AP ∼
= BP ∼
= CP and that ` is perpendicular to each of
these line segments. Thus, the triangles 4ABP and 4ACP 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, 4ABP ∼
6 4ACP because the length of
=
AB is strictly less than the length of AC.
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