Proof: By Euclid
(related to Proposition: 1.02: Constructing a Segment Equal to an Arbitrary Segment)
- For let the straight line $AB$ have been joined from point $A$ to point $B$ [Post. 1] , and let the equilateral triangle $DAB$ have been been constructed upon it [Prop. 1.1].
- And let the straight lines $AE$ and $BF$ have been produced in a straight line with $DA$ and $DB$ (respectively) [Post. 2] .
- And let the circle $CGH$ with center $B$ and radius $BC$ have been drawn [Post. 3] , and again let the circle $GKL$ with center $D$ and radius $DG$ have been drawn [Post. 3] .
- Therefore, since the point $B$ is the center of (the circle) $CGH$, $BC$ is equal to $BG$ [Def. 1.15] .
- Again, since the point $D$ is the center of the circle $GKL$, $DL$ is equal to $DG$ [Def. 1.15] .
- And within these, $DA$ is equal to $DB$.
- Thus, the remainder $AL$ is equal to the remainder $BG$ [C.N. 3] .
- But $BC$ was also shown (to be) equal to $BG$.
- Thus, $AL$ and $BC$ are each equal to $BG$.
- But things equal to the same thing are also equal to one another [C.N. 1] .
- Thus, $AL$ is also equal to $BC$.
- Thus, the straight line $AL$, equal to the given straight line $BC$, has been placed at the given point $A$.
- (Which is) the very thing it was required to do.
fn1. This proposition admits of a number of different cases, depending on the relative positions of the point $A$ and the line $BC$. In such situations, Euclid invariably only considers one particular case - usually, the most difficult - and leaves the remaining cases as exercises for the reader (translator's note).
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"