Proof: By Euclid

So let a third (part) have been prescribed.
And let some straight line $AC$ have been drawn from (point) $A$, encompassing a random angle with $AB$.
And let a random point $D$ have been taken on $AC$.
And let $DE$ and $EC$ be made equal to $AD$ [Prop. 1.3].
And let $BC$ have been joined.
And let $DF$ have been drawn through $D$ parallel to it [Prop. 1.31].
Therefore, since $FD$ has been drawn parallel to one of the sides, $BC$, of triangle $ABC$, then, proportionally, as $CD$ is to $DA$, so $BF$ (is) to $FA$ [Prop. 6.2].
And $CD$ (is) double $DA$.
Thus, $BF$ (is) also double $FA$.
Thus, $BA$ (is) triple $AF$.
Thus, the prescribed third part, $AF$, has been cut off from the given straight line, $AB$.
(Which is) the very thing it was required to do.
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