show that there is a magnitude p such that the area of a circle of radius r is to the area of a square of side r as p is to 1. a:b=d:d

Proof is based on:

The proof depends on three earlier results in the Elements:

- Given two magnitudes AB > CD, if from AB a quantity more than half is removed, and from the remainder a a quantity more than half is removed, and so on, eventually what is left will be less than CD.

ProofOne of the axiomatic properties of magnitudes is that there is a positive integer n such that n CD > AB, and a property of intgers is that there is a positive integer m such that 2m > n. Thus there is an m such that CD : AB > 1:2m. Then the remainder after the m -th cut off is < 1/(2m)AB < CD.

- Let P and Q be regular similar polygons with correspnding side lengths p and q . Then Area P : Area Q = p2 : q2.

Since this is true for similar triangles, by the construction of Thales, it is true for all similar polygons by decomposition.

- The
Law of Trichotomy(not explicitly stated, but used throughout): If x and y are magnitudes, then x < y or x = y or x > y . To us, this is an axiom of the real numbers; to the Greeks it was an unstated axiom of magnitudes.

CirclesC and D are to one another as thesquareson their diameters c and d .

Either Area C : Area D = c2 : d2 or < c2 : d2 or > c2 : d2.

Suppose Area C : Area D < c2 : d2, so Area D > Area C.(d2/c2) . Let S = Area C.(d2/c2) , so S < Area D .

By Proposition 1 of Book X, we can continually remove from Area D larger and larger polygons inscribed in D until we get a polygon P such that S < Area P < Area D, so Area P > Area C. d2/c2.

Now inscribe in C a polygon Q similar to P so Area Q < Area C and Area Q : Area P = c2:d2. Thus Area P = Area Q . (d2/c2) < Area C. (d2/c2), a contradiction.

A similar contradiction arises if Area C : Area D > c2 : d2 and we inscribe a polygon in C .