Proof: By Euclid

For let ( $BA$ and $AC$ ) have been produced to points $D$ and $E$ (respectively), and let $BD$ be made equal to $AC$ [Prop. 1.3].
And let $BC$ have been joined.
And let $DE$ have been drawn through (point) $D$ parallel to it [Prop. 1.31].
Therefore, since $BC$ has been drawn parallel to one of the sides $DE$ of triangle $ADE$ , proportionally, as $AB$ is to $BD$ , so $AC$ (is) to $CE$ [Prop. 6.2].
And $BD$ (is) equal to $AC$ .
Thus, as $AB$ is to $AC$ , so $AC$ (is) to $CE$ .
Thus, a third (straight line), $CE$ , has been found (which is) proportional to the two given straight lines, $AB$ and $AC$ .
(Which is) the very thing it was required to do.¹

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It has been shown that it is possible to construct a segment with a squared ratio to two segments with a given ratio. This construction can be repeated to construct a segment with a cubed ratio or an arbitrary continued proportion (editor's note). ↩