Similar Figures on Proportional Straight Lines

Theorem

 * 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.

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 the porism to Ratio of Areas of Similar Triangles $AB : O = KAP : LCD$ and $EF : P = MF : NH$.

Therefore from Equality of Ratios is Transitive $KAB : LCD = MF : NH$.


 * Euclid-VI-22.png

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 $AB : CD = EF : GH$.