On 2/14/2014 6:00 PM, Ken Quirici wrote: > According to Wikipedia, Book 1 Proposition 1 (constructing equilateral triangle on given line segment) > is missing a premise - that if two different circles share a radius (that is, a line segment connecting > their centers is a radius for both circles). > > Could somebody provide a proof of this premise? > > Regards, > > Ken >
The circle fills space uniformly with the least perimeter for area so they are the same. It is half-way then for the equilateral, because the angles are equi-angular, in the evolution of angles, they are equal.
Sorry, this is a rather poor geometric construction, I'd be most surprised to hear of something like that in Euclid.
Then, it would because the radii at a distance from the line with the parallel line, with the adjustable compass from drawing the parallel lines, would form the triangles with that height that form the equilateral triangle (that the side is equal, to find then the parallel line at the height of the equilateral triangle. This is finding the center from shifting that up then finding the mid-point of that, bisecting the segment with the compass and edge, then halving and doubling, for the projection of the triangle that is of the parallel lines, to the orthogonal parallel distance that is of the third side of the equilateral triangle, through that.