Proof: By Euclid

Let the two straight lines $DE$ and $DF$ be set out encompassing the [random] angle $EDF$.
And let $DG$ be made equal to $A$, and $GE$ to $B$, and, further, $DH$ to $C$ [Prop. 1.3].
And $GH$ being joined, let $EF$ have been drawn through (point) $E$ parallel to it [Prop. 1.31].
Therefore, since $GH$ has been drawn parallel to one of the sides $EF$ of triangle $DEF$, thus as $DG$ is to $GE$, so $DH$ (is) to $HF$ [Prop. 6.2].
And $DG$ (is) equal to $A$, and $GE$ to $B$, and $DH$ to $C$.
Thus, as $A$ is to $B$, so $C$ (is) to $HF$.
Thus, a fourth (straight line), $HF$, has been found (which is) proportional to the three given straight lines, $A$, $B$, and $C$.
(Which is) the very thing it was required to do.
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