Euclid Book 5 Proposition 21
Statement
Given two groups of three magnitudes (AB, CD, and EF; GH, IJ, and KL), if the first magnitude is to the second as the fifth is to the sixth (), the second is to the third as the fourth is to the fifth (), and the first magnitude is greater than or equal to the third (AB≥EF), then the fourth magnitude is also greater than or equal to the sixth (GH≥KL).
AB
CD
IJ
KL
CD
EF
GH
IJ
Computational Explanation
GeometricScene{{A.,B.,C.,D.,E.,F.,G.,H.,I.,J.,K.,L.},{a.,b.,c.,d.,e.,f.}},Style[Line[{A.,B.}],Red],Style[Line[{C.,D.}],Red],Style[Line[{E.,F.}],Red],Style[Line[{G.,H.}],Blue],Style[Line[{I.,J.}],Blue],Style[Line[{K.,L.}],Blue],EuclideanDistance[A.,B.]a.,EuclideanDistance[C.,D.]b.,EuclideanDistance[E.,F.]c.,EuclideanDistance[G.,H.]d.,EuclideanDistance[I.,J.]e.,EuclideanDistance[K.,L.]f.,,,a.≥c.,{EuclideanDistance[G.,H.]≥EuclideanDistance[K.,L.],d.≥f.}
a.
b.
e.
f.
b.
c.
d.
e.
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Explanations
Let there be three magnitudes , , , and others , , equal to them in multitude, which taken two and two are in the same ratio, and let the proportion of them be perturbed, so that, as is to , so is to , and, as is to , so is to , and let be greater than ex aequali; I say that will also be greater than ; if is equal to , equal; and if less, less.
For, since is greater than , and is some other magnitude, therefore has to a greater ratio than has to .
But, as is to , so is to, and, as is to , inversely, so is to . Therefore also has to a greater ratio than has to .
But that to which the same has a greater ratio is less; therefore is less than ; therefore is greater than .
Similarly we can prove that, if be equal to , will also be equal to ; and if less, less.
A
B
C
D
E
F
A
B
E
F
B
C
D
E
A
C
D
F
A
C
For, since
A
C
B
A
B
C
B
But, as
A
B
E
F
C
B
E
D
E
F
E
D
But that to which the same has a greater ratio is less; therefore
F
D
D
F
Similarly we can prove that, if
A
C
D
F