Proof: By Euclid
(related to Proposition: 6.30: Construction of the Inverse Golden Section)
- Let the square BC have been described on AB [Prop. 1.46], and let the parallelogram CD, equal to BC, have been applied to AC, overshooting by the figure AD (which is) similar to BC [Prop. 6.29].
- And BC is a square.
- Thus, AD is also a square.
- And since BC is equal to CD, let (rectangle) CE have been subtracted from both.
- Thus, the remaining (rectangle) BF is equal to the remaining (square) AD.
- And it is also equiangular to it.
- Thus, the sides of BF and AD about the equal angles are reciprocally proportional [Prop. 6.14].
- Thus, as FE is to ED, so AE (is) to EB.
- And FE (is) equal to AB, and ED to AE.
- Thus, as BA is to AE, so AE (is) to EB.
- And AB (is) greater than AE.
- Thus, AE (is) also greater than EB [Prop. 5.14].
- Thus, the straight line AB has been cut in extreme and mean ratio at E, and AE is its greater piece.
- (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"