# Similar Figures on Proportional Straight Lines

## Theorem

In the words of Euclid:

*If four straight lines be proportional, the rectilineal figures similar and similarly described upon them will also be proportional; and if the rectilineal figures similar and similarly described upon them be proportional, the straight lines will themselves be proportional.*

(*The Elements*: Book $\text{VI}$: Proposition $22$)

## Proof

Let the four straight lines $AB, CD, EF, GH$ be proportional.

That is:

- $AB : CD = EF : GH$

Let there be described on $AB$ and $CD$ the similar and similarly situated rectilineal figures $KAB$ and $LCD$.

Let there be described on $EF$ and $GH$ the similar and similarly situated rectilineal figures $MF$ and $NH$.

We need to show that:

- $KAB : LCD = MF : NH$

By Construction of Third Proportional Straight Line, let there be taken a third proportional $O$ to $AB, CD$, and a third proportional $P$ to $EF, GH$.

We have that:

- $AB : CD = EF : GH$, and $CD : O = GH : P$

So from Equality of Ratios Ex Aequali:

- $AB : O = EF : P$

But from Porism to Proposition $19$ of Book $\text{VI} $: Ratio of Areas of Similar Triangles:

- $AB : O = KAP : LCD$

and:

- $EF : P = MF : NH$

Therefore from Equality of Ratios is Transitive:

- $KAB : LCD = MF : NH$

Now suppose $MF : NH = KAB : LCD$.

We need to show that:

- $AB : CD = EF : GH$

Suppose, with a view to obtaining a contradiction, that:

- $EF : GH \ne AB : CD$

Instead, by Construction of Fourth Proportional Straight Line, let:

- $EF : QR = AB : CD$

On $QR$, by Construction of Similar Polygon, let the rectilineal figures $SR$ be described similar and similarly situated to either $MF$ or $NH$.

We have that:

- $AB : CD = EF : QR$

We also have that there have been described on $AB$ and $CD$ the similar and similarly situated rectilineal figures $KAB$ and $LCD$

We also have that there have been on $EF$ and $GH$ the similar and similarly situated rectilineal figures $MF$ and $NH$.

From Equality of Ratios is Transitive:

- $MF : SF = MF : NH$

Therefore $MF$ has the same ratio to each of the figures $NH, SR$.

Therefore, by Magnitudes with Same Ratios are Equal:

- $NH = SR$

But it is also similar and similarly situated to it.

Therefore:

- $GH = QR$

But we have that $AB : CD = EF : QR$ while $QR = GH$.

Therefore:

- $AB : CD = EF : GH$

$\blacksquare$

## Historical Note

This proof is Proposition $22$ of Book $\text{VI}$ 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 $\text{VI}$. Propositions