Proof: By Euclid
(related to Proposition: 6.09: Construction of Part of Line)
- 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.
Thank you to the contributors under CC BY-SA 4.0!
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"