Euclid Book 5 Proposition 6
Statement
If two magnitudes are equimultiples of two other magnitudes (AB5EF, CD5GH), and two magnitudes subtracted from the former are equimultiples of the latter (AI3EF, CJ3GH), then the remainders are equimultiples of them as well (IB2EF, JD2GH).
Computational Explanation
GeometricSceneA.,B.,C.,D.,E.,F.,G.,H.,I.,J.,a.,b.,Line[A.,I.,B.],LineC.,J.,D.,StyleLine[E.,F.],Red,StyleLineG.,H.,Blue,StyleLine[A.,I.],Red,StyleLineC.,J.,Blue,EuclideanDistance[A.,B.]5EuclideanDistance[E.,F.]5a.,EuclideanDistanceC.,D.5EuclideanDistanceG.,H.5b.,EuclideanDistance[A.,I.]3EuclideanDistance[E.,F.]3a.,EuclideanDistanceC.,J.3EuclideanDistanceG.,H.3b.,EuclideanDistance[I.,B.]2EuclideanDistance[E.,F.],EuclideanDistanceJ.,D.2EuclideanDistanceG.,H.,5a.-3a.(5-3)a.,5b.-3b.(5-3)b.
| ||
 |
Explanations
For let two magnitudes , be equimultiples of two magnitudes , , and let , subtracted from them be equimultiples of the same two , ; I say that the remainders also, , , are either equal to , or equimultiples of them.
For, first, let be equal to ; I say that is also equal to .
For let be made equal to .
Since is the same multiple of that is of , while is equal to and 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 .
Since then each of the magnitudes, is the same multiple of , therefore is equal to .
Let be subtracted from each; therefore the remainder is equal to the remainder . But is equal to ; therefore is also equal to .
Hence, if is equal to , is also equal to .
Similarly we can prove that, even if be a multiple of , is also the same multiple of .
AB
CD
E
F
AG
CH
E
F
GB
HD
E
F
For, first, let
GB
E
HD
F
For let
CK
F
Since
AG
E
CH
F
GB
E
KC
F
AB
E
KH
F
But, by hypothesis,
AB
E
CD
F
KH
F
CD
F
Since then each of the magnitudes
KH
CD
F
KH
CD
Let
CH
KC
HD
F
KC
HD
F
Hence, if
GB
E
HD
F
Similarly we can prove that, even if
GB
E
HD
F