Euclid Book 5 Proposition 1
Statement
If a first magnitude is the same multiple of a second that a third is of a fourth (AB2CD, EF2GH), then the sum of the first and third also is the same multiple of the sum of the second and fourth (AB+EF2(CD+GH)) that the third is of the fourth.
Computational Explanation
GeometricScene[{{A.,B.,C.,D.,E.,F.,G.,H.},{a.,b.}},{Style[Line[{A.,B.}],Blue],Style[Line[{C.,D.}],Blue],Style[Line[{E.,F.}],Red],Style[Line[{G.,H.}],Red],EuclideanDistance[A.,B.]2EuclideanDistance[C.,D.]2a.,EuclideanDistance[E.,F.]2EuclideanDistance[G.,H.]2b.},{EuclideanDistance[A.,B.]+EuclideanDistance[E.,F.]2(EuclideanDistance[C.,D.]+EuclideanDistance[G.,H.]),2a.+2b.2(a.+b.)}]
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Explanations
Let any number of magnitudes whatever , be respectively equimultiples of any magnitudes , equal in multitude; I say that, whatever multiple is of , that multiple will , also be of , . For, since is the same multiple of that is of , as many magnitudes as there are in equal to , so many also are there in equal to . Let be divided into the magnitudes , equal to , and into CH, HD equal to ; then the multitude of the magnitudes , will be equal to the multitude of the magnitudes , .
Now, since is equal to , and to , therefore is equal to , and , to , .
For the same reason is equal to , and , to , ; therefore, as many magnitudes as there are in equal to , so many also are there in , equal to , ; therefore, whatever multiple is of , that multiple will , also be of , .
AB
CD
E
F
AB
E
AB
CD
E
F
AB
E
CD
F
AB
E
CD
F
AB
AG
GB
E
CD
F
AG
GB
CH
HD
Now, since
AG
E
CH
F
AG
E
AG
CH
E
F
For the same reason
GB
E
GB
HD
E
F
AB
E
AB
CD
E
F
AB
E
AB
CD
E
F