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.
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes