# Equality of Ratios is Transitive

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## Contents

## Theorem

In the words of Euclid:

(*The Elements*: Book $\text{V}$: Proposition $11$)

That is:

- $A : B = C : D, C : D = E : F \implies A : B = E : F$

## Proof

Let $A : B = C : D$ and $C : D = E : F$.

Of $A, C, E$ let equimultiples $G, H, K$ be taken.

Of $B, D, F$ let other arbitrary equimultiples $L, M, N$ be taken.

We have that:

- $A : B = C : D$
- $G, H$ are equimultiples of $A, C$
- $L, M$ are equimultiples of $B, D$

So:

- $G > L \implies H > M$
- $G = L \implies H = M$
- $G < L \implies H < M$

Also, we have that:

- $C : D = E : F$
- $H, K$ are equimultiples of $C, E$
- $M, N$ are equimultiples of $D, F$

So:

- $H > M \implies K > N$
- $H = M \implies K = N$
- $H < M \implies K < N$

Also, we have that:

- $G, K$ are equimultiples of $A, E$
- $L, N$ are equimultiples of $B, F$

Therefore $A : B = E : F$.

$\blacksquare$

## Historical Note

This theorem is Proposition $11$ of Book $\text{V}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 2*(2nd ed.) ... (previous) ... (next): Book $\text{V}$. Propositions