Quote: geometric construction of equilateral triangles; with proof

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constructing proof and program together


BOOK 1. PROPOSITIONS Proposition 1 On a given finite straight line to construct an equilateral triangle. ... With centre A and distance AB let the circle BCD be described; [Post. 3] again, with centre B and distance BA let the circle ACE be described; [Post. 3] and from the point C, in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined. [Post. 1] Now, since the pint A is the centre of the circle CDB, AC is equal to AB. [Def. 15] Again, since the point B is the centre of the circle CAE, BC is equal to BA; [Def. 15] But CA was also proved equal to AB; therefore each of the straight lines CA, CB is equal to AB. And things which are equal to the same thing are also equal to one another; therefore CA is also equal to CB. [C.N. 1] Therefore the three straight lines CA, AB, BC are equal to one another. [p. 3] Therefore the triangle ABC is equilateral; and it has been constructed on the given finite straight line AB. (Being) what is was required to do.   Google-1   Google-2

Published before 1923

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