Euclid Book 5 Proposition 5
Statement
If a first magnitude is the same multiple of a second magnitude that a subtracted part of the first is of a subtracted part of the second (AB3CD, AE3CF), then the remainder of the first magnitude also is the same multiple of the remainder of the second that the first whole is of the second whole (EB3FD).
Computational Explanation
GeometricSceneA.,B.,C.,D.,E.,F.,a.,b.,Line[A.,E.,B.],StyleLine[A.,E.],Red,StyleLine[E.,B.],Blue,LineC.,F.,D.,StyleLineC.,F.,Red,StyleLine[F.,D.],Blue,EuclideanDistance[A.,B.]3EuclideanDistanceC.,D.3a.,EuclideanDistance[A.,E.]3EuclideanDistanceC.,F.3b.,EuclideanDistance[E.,B.]3EuclideanDistance[F.,D.],3a.-3b.3a.-b.
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Explanations
For let the magnitude be the same multiple of the magnitude that the part subtracted is of the part subtracted; I say that the remainder is also the same multiple of the remainder that the whole is of the whole .
For, whatever multiple is of , let be made that multiple of . Then, since is the same multiple of that is of , therefore is the same multiple of that is of .
But, by the assumption, is the same multiple of that is of .
Therefore is the same multiple of each of the magnitudes , ; therefore is equal to .
Let be subtracted from each; therefore the remainder is equal to the remainder .
And, since is the same multiple of that is of , and is equal to , therefore is the same multiple of that is of .
But, by hypothesis, is the same multiple of that is of ; therefore is the same multiple of that is of .
That is, the remainder will be the same multiple of the remainder that the whole is of the whole .
AB
CD
AE
CF
EB
FD
AB
CD
For, whatever multiple
AE
CF
EB
CG
AE
CF
EB
GC
AE
CF
AB
GF
But, by the assumption,
AE
CF
AB
CD
Therefore
AB
GF
CD
GF
CD
Let
CF
GC
FD
And, since
AE
CF
EB
GC
GC
DF
AE
CF
EB
FD
But, by hypothesis,
AE
CF
AB
CD
EB
FD
AB
CD
That is, the remainder
EB
FD
AB
CD