Proof: By Euclid
(related to Proposition: 6.11: Construction of Segment in Squared Ratio)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes