Euclid Book 5 Proposition 20
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 fourth is to the fifth (), the second is to the third as the fifth is to the sixth (), 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
GH
IJ
CD
EF
IJ
KL
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.
d.
e.
b.
c.
e.
f.
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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, 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, and the greater has to the same a greater ratio than the less has, 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 has also to a greater ratio than has to .
But, of magnitudes which have a ratio to the same, that which has a greater ratio is greater; 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
D
E
B
C
E
F
A
C
D
F
A
C
For, since
A
C
B
A
B
C
B
But, as
A
B
D
E
C
B
F
E
D
E
F
E
But, of magnitudes which have a ratio to the same, that which has a greater ratio is greater; therefore
D
F
Similarly we can prove that, if
A
C
D
F