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