Logical Structure of the Argument
Heuristic Version of the Argument
Theorem (Elements xii 1): If similar polygons are inscribed
in circles, their ratios are as the squares of the diameters of the circles.
Theorem 2: Circles are to one another as the squares on the diameters (trans. T.L. Heath, with modifications).
p. 1
Let there be circles ABCD, EFGH, and their diameters BD, FH; I say that, as the circle ABCD is to the circle EFGH, so is the square on BD to the square on FH.  circle ABCD : circle EFGH = BD^{2} : FH^{2} 
p. 2
Basic Assumption of 4th proportional: where a : b and c : d are ratios, (a : b c : d) => x(a : b = c : x and x > d or x < d)
For,
if the square on BD is not to the square on FH as the circle ABCD is to
the circle EFGH, then, so will the circle ABCD be either to some less area
than the circle EFGH, or to a greater.
First, let it be in that ratio to a less area S. 
Suppose not:
Then, circle ABCD : S = BD^{2} : FH^{2} Where S < circle EFGH
Suppose S < circle EFGH. 
p. 3
Strategy: Take away areas from the circle EFGH or what is left of the circle until we have an area less than circle EFGH  S.
By Elements X 1, each removal must be more than half of the remainder.
The circle less the areas left forms a polygon (or, as we might put it, the sum of the areas taken away). We show that this polygon is less than S.
Let the square EFGH be inscribed in the circle EFGH;  Step 1: We inscribe a square and show that it is more than half the circle and so takes away more than half the circle. 
p. 4
then the inscribed square is greater than the half of the circle EFGH, inasmuch as, if through the points E, F, G, H we draw tangents to the circle, the square EFGH is half the square circumscribed about the circle, and the circle is less than the circumscribed square; hence the inscribed square EFGH is greater than the half of the circle EFGH.  Square EFGH = 1/2 circumscribed square
circumscribed square > circle EFGH Hence,

p. 5
Let the circumferences EF, FG, GH, HE be bisected at the points K, L, M, N, and let EK, KF, FL, LG, GM, MH, HN, NE be joined;  Step n (actually step 2, but generalizable): We inscribe
a 2^{n+1}gon and show that the triangles added to the previous
2^{n}gon to construct the 2^{n+1}gon are more than half
what remains of the circle and so take away more than half the remainder.
Bisect the arcs about the squares and form the triangles. 
p. 6
therefore each of the triangles EKF, FLG, GMH, HNE is also greater than the half of the segment of the circle about it, inasmuch as, if through the points K, L, M, N we draw tangents to the circle and complete the parallelograms on the straight lines EF, FG, GH, HE, each of the triangles EKF, FLG, GMH, HNE will be half of the parallelogram about it, while the segment about it is less than the parallelogram; hence each of the triangles EKF, FLG, GMH, HNE is greater than the half of the segment of the circle about it.  each triangle EKF = 1/2 the rectangle about the arc.
the rectangle > the arc about the base of the triangle the each triangle > 1/2 the rectangle about the arc. 
p. 7
Thus, by bisecting the remaining circumferences and joining straight lines, and by doing this continually, we shall leave some segments of the circle which will be less than the excess by which the circle EFGH exceeds the area S. For it was proved in the first theorem of the tenth book that, if two unequal magnitudes be set out, and if from the greater there by subtracted a magnitude greater than the half, and from that which is left a greater than the half, and if this be done continually, there will be left some magnitude which will be less than the lesser magnitude set out.  By Euclid, Elements X 2 (quoted in the text), there will be a step n such that circle EFGH  the 2^{n+1}gon < circle EFGH  S. 
p. 8
Let segments be left such as described, and let the segments of the circle EFGH on EK, KF, FL, LG, GM, MH, HN, NE be less than the excess by which the circle EFGH exceeds the area S. Therefore the remainder, the polygon EKFLGMHN, is greater than the area S.  Since
the circle EFGH  the 2^{n+1}gon < circle EFGH  S it follows that the 2^{n+1}gon > S. 
p. 9
Let
there be inscribed, also, in the circle ABCD the polygon AOBPCQDR similar
to the polygon EKFLGMHN; therefore, as the square on BD is to the square
on FH, so is the polygon AOBPCQDR to the polygon EKFLGMHN.
But, as the square on BD is to the square on FH, so also is the circle ABCD to the area S; therefore also, as the circle ABCD is to the area S, so is the polygon AOBPCQDR to the polygon EKFLGMHN. But the circle ABCD is greater than the polygon inscribed in it; therefore the area S is also greater than the polygon EKFLGMHN. But it is also less which is impossible. Therefore, as the square on BD is to the square on FH, so is not the circle ABCD to any area less than the circle EFGH. 
Construct a similar 2^{n+1}gon in circle ABCD.
The 2^{n+1}gon in circle ABCD : the 2^{n+1}gon in circle EFGH = BD^{2} : FH^{2} (by Theorem XII 1) BD^{2} : FH^{2} = circle ABCD : S (by the Hypothesis). Hence,
Hence,
circle ABCD > the 2^{n+1}gon in circle ABCD Hence, S > the 2^{n+1}gon in circle EFGH. Hence, S is larger and less than the 2^{n+1}gon in circle EFGH, which is impossible. Hence, S is not less than circle EFGH (the hypothesis). 
p. 10
Similarly we can prove that neither is the circle EFGH to any area less than the circle ABCD as the square on FH is to the square on BD.  By the same argument,
circle EFGH : S * FH^{2} : BD^{2} if S < circle ABCD 
p. 11
p. 12
Therefore, circles are to one another as the squares on the diameters, which it was necessary to prove (Q.E.D.)  Hence, it is not the case that S > circle EFGH nor S < circle EFGH. Hence, S = EFGH. 
p. 13
Lemma
I say that, the area S being greater than the circle EFGH, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD. For let it be contrived that, as the area S is to the circle ABCD, so is the circle EFGH to the area T. I say that the area T is less than the circle ABCD. For since, as the area S is to the circle ABCD, so is the circle EFGH to the area T, therefore, alternately, as the area S is to the circle EFGH, so is the circle ABCD to the area T. But the area S is greater than the circle EFGH; therefore the circle ABCD is also greater than the area T. Hence, as the area S is to the circle ABCD, so is the circle EFGH to some area less than the circle ABCD, which it was necessary to prove (Q.E.D.) 
The lemma is argued above. 
p. 14
The argument uses a principle of the existence of the fourth proportional and an assumption of connection:We can separate out the two assumptions built into this assumption as
follows:
Existence of the fourth proportional: a : b is a ratio and c
a continuous quantity x(a
: b = c : x)
Connection: c : d and c : x are ratios => (x = d
x > d x < d)
The proof shows that: x(a : b = c : x and (x d or x d)) and so infers that $x(a : b = c : x and x = d)
Use of Elements X 1: let A > B and we construct a series, A, A_{1}, ..., A_{n}, ..., such that A_{n+1} = A_{n}  X_{n+1}, where X_{n} > 1/2 A_{n}. Then there is an A_{n} such that A_{n} < B.
By the assumption of the 4th proportional and the supposition that a : b c : d, we suppose that a : s = c : d, where s > b or s < b.
Case 1a: s < b:
We suppose that a : s = c : d and s < b. By construction and
Elements
X 1, we find r such that r < bs. Hence br
> s.
By construction we show that there is a p < a such that p : br
= c : d, and by the hypothesis c : d = a : s.
Hence p : br = a : s. By alternando, a : p = s : br.
Since a > p, s > br, a contradiction.
Hence, s b.
Case 1b: The same argument will show that if (b : a d : c) and we hypothesize s such that b : s = d : c, the assumption that s < a will likewise lead to contradiction.
Case 2: s > b:
We suppose that a : s = c : d and s > b. By the assumption of
the 4th proportional, we find t such that a : s = t : b = c : d or s :
a = b : t = d : c. By alternando, a : t = s : b. Since s >
b (by hypothesis), a > t. Hence, b : t = d : c, where t < a.
This reduces Case 2 to Case 1b.
Some salient features: