# Proof: By Euclid

• Let $AB$, $CD$, $EF$, and $GH$ be four proportional straight lines, (such that) as $AB$ (is) to $CD$, so $EF$ (is) to $GH$.
• And let the similar, and similarly laid out, rectilinear figures $KAB$ and $LCD$ have been described on $AB$ and $CD$ (respectively), and the similar, and similarly laid out, rectilinear figures $MF$ and $NH$ on $EF$ and $GH$ (respectively).
• I say that as $KAB$ is to $LCD$, so $MF$ (is) to $NH$.
• And so let $KAB$ be to $LCD$, as $MF$ (is) to $NH$.
• I say also that as $AB$ is to $CD$, so $EF$ (is) to $GH$.
• For if as $AB$ is to $CD$, so $EF$ (is) not to $GH$, let $AB$ be to $CD$, as $EF$ (is) to $QR$ [Prop. 6.12].
• And let the rectilinear figure $SR$, similar, and similarly laid down, to either of $MF$ or $NH$, have been described on $QR$ [Prop. 6.18], [Prop. 6.21].
• Therefore, since as $AB$ is to $CD$, so $EF$ (is) to $QR$, and the similar, and similarly laid out, (rectilinear figures) $KAB$ and $LCD$ have been described on $AB$ and $CD$ (respectively), and the similar, and similarly laid out, (rectilinear figures) $MF$ and $SR$ on $EF$ and $QR$ (resespectively), thus as $KAB$ is to $LCD$, so $MF$ (is) to $SR$ (see above).
• And it was also assumed that as $KAB$ (is) to $LCD$, so $MF$ (is) to $NH$.
• Thus, also, as $MF$ (is) to $SR$, so $MF$ (is) to $NH$ [Prop. 5.11].
• Thus, $MF$ has the same ratio to each of $NH$ and $SR$.
• Thus, $NH$ is equal to $SR$ [Prop. 5.9].
• And it is also similar, and similarly laid out, to it.
• Thus, $GH$ (is) equal to $QR$.1
• And since $AB$ is to $CD$, as $EF$ (is) to $QR$, and $QR$ (is) equal to $GH$, thus as $AB$ is to $CD$, so $EF$ (is) to $GH$.
• Thus, if four straight lines are proportional, then similar, and similarly described, rectilinear figures (drawn) on them will also be proportional.
• And if similar, and similarly described, rectilinear figures (drawn) on them are proportional then the straight lines themselves will also be proportional.
• (Which is) the very thing it was required to show.

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