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Euclid Book 5 Proposition 17

Euclid Book 5 Proposition 17

Statement

If a whole is to a subtracted part as another whole is to another subtracted part (

AB

EB



CD

FD

), then the remainder of the former is to the subtracted part as the remainder of the latter is to the subtracted part (

AE

EB



CF

FD

).

Computational Explanation

GeometricScene{{A.,B.,C.,D.,E.,F.},{a.,b.,c.,d.}},Line[{A.,E.,B.}],Line[{C.,F.,D.}],Style[Line[{A.,E.}],Red],Style[Line[{E.,B.}],Pink],Style[Line[{C.,F.}],Blue],Style[Line[{F.,D.}],Cyan],EuclideanDistance[A.,E.]a.,EuclideanDistance[E.,B.]b.,

EuclideanDistance[A.,B.]

EuclideanDistance[E.,B.]



a.+b.

b.

,EuclideanDistance[C.,F.]c.,EuclideanDistance[F.,D.]d.,

EuclideanDistance[C.,D.]

EuclideanDistance[F.,D.]



c.+d.

d.

,

a.+b.

b.



c.+d.

d.

,

EuclideanDistance[A.,E.]

EuclideanDistance[E.,B.]



EuclideanDistance[C.,F.]

EuclideanDistance[F.,D.]

,

a.

b.



c.

d.



Conclusion

EuclideanDistance[A.,E.]

EuclideanDistance[E.,B.]



EuclideanDistance[C.,F.]

EuclideanDistance[F.,D.]

Explanations

Let

AB

,

BE

,

CD

,

DF

be magnitudes proportional componendo, so that, as

AB

is to

BE

, so is

CD

to

DF

; I say that they will also be proportional separando, that is, as

AE

is to

EB

, so is

CF

to

DF

.

For of

AE

,

EB

,

CF

,

FD

let equimultiples

GH

,

HK

,

LM

,

MN

be taken, and of

EB

,

FD

other, chance, equimultiples,

KO

,

NP

.

Then, since

GH

is the same multiple of

AE

that

HK

is of

EB

, therefore

GH

is the same multiple of

AE

that

GK

is of

AB

.

But

GH

is the same multiple of

AE

that

LM

is of

CF

; therefore

GK

is the same multiple of

AB

that

LM

is of

CF

.

Again, since

LM

is the same multiple of

CF

that

MN

is of

FD

, therefore

LM

is the same multiple of

CF

that

LN

is of

CD

.

But

LM

was the same multiple of

CF

that

GK

is of

AB

; therefore

GK

is the same multiple of

AB

that

LN

is of

CD

.

Therefore

GK

,

LN

are equimultiples of

AB

,

CD

.

Again, since

HK

is the same multiple of

EB

that

MN

is of

FD

, and

KO

is also the same multiple of

EB

that

NP

is of

FD

, therefore the sum

HO

is also the same multiple of

EB

that

MP

is of

FD

.

And, since, as

AB

is to

BE

, so is

CD

to

DF

, and of

AB

,

CD

equimultiples

GK

,

LN

have been taken, and of

EB

,

FD

equimultiples

HO

,

MP

, therefore, if

GK

is in excess of

HO

,

LN

is also in excess of

MP

, if equal, equal, and if less, less.

Let

GK

be in excess of

HO

; then, if

HK

be subtracted from each,

GH

is also in excess of

KO

.

But we saw that, if

GK

was in excess of

HO

,

LN

was also in excess of

MP

; therefore

LN

is also in excess of

MP

, and, if

MN

be subtracted from each,

LM

is also in excess of

NP

; so that, if

GH

is in excess of

KO

,

LM

is also in excess of

NP

.

Similarly we can prove that, if

GH

be equal to

KO

,

LM

will also be equal to

NP

, and if less, less.

And

GH

,

LM

are equimultiples of

AE

,

CF

, while

KO

,

NP

are other, chance, equimultiples of

EB

,

FD

; therefore, as

AE

is to

EB

, so is

CF

to

FD

.

Classes

Euclid's Elements
Theorems
Geometric Algebra
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