The theorem discussed is basic to the proofs of Archimedes, On Conoids and Spheroids, and is used in the Method.

Theorem and Proof with Modern Notation
Theorem:  let there be two series of equal size:

A1, .., An, ...
B1, .., Bn, ...
and for corresponding magnitudes, Ai, Aj to Bi, Bj,
a)  Ai : Aj = Bi : Bj
(Note that there must be a natural pairing if the sets are infinite)

and there are series of magnitudes:

C1, .., Cn, ...
D1, .., Dn, ...
where
b)  Ai : Ci = Bi : Di
then
(A1+ ... + An+ ...) : (C1+ ... + Cn+ ...) = (B1+ ... + Bn+ ...) : (D1+ ... + Dn+ ....)

The proof below frequently uses the rule, ex aequali, namely,   a : b = e : f and b : c = f : g => a : c = e : g.

Proof:

 1.    C1 : A1 = D1 : B1 Assumption (b) and a : b = c : d => b : a = d : c 2.    A1 : A2 = B1 : B2 Assumption (a) 3.     A2  : C2 : = B2 : D2 Assumption (b) 4.     C1 : C2 = D1 : D2 Two applications of Ex aequali (1 and 2, then the result and 3):  a : b = c : d and b : e = d : f => a : e = c : f. 5.     Ci : Cj = Di : Dj By repitition of the reasoning in 1-4 6.     (A1+ ... + An) : A1 = (B1+ ... + Bn) : B1 Since by assumption (a) Ai : A1 = Bi : B1 and Aj : A1 = Bj : B1, it follows that (Ai + Aj) : A1 = (Bi + Bj) : B1.  Repeat through the whole set. 7.    A1 : C1 = B1 : D1 Assumption 8.    C1 : (C1+ ... + Cn)  = D1 : (D1+ ... + Dn) By (5) and the same reasoning as (6) above 9.     (A1+ ... + An) : (C1+ ... + Cn)                      = (B1+ ... + Bn) : (D1+ ... + Dn) Ex aequali (7 and 8)

Translation of On Conoids and Spheroids 1
If any magnitudes, however many in multitude, have the same ratio when taken pairwise as other magnitudes equal in number, when similarly arranged, and the first magnitudes are in some ratio to other magnitudes (whether all or some of them), and the second magnitudes are correspondingly to other magnitudes in the same ratio [as the first magnitudes to the members of the third group], then all the first magnitudes have the same ratio to all those picked out for them, which all the second have to all those picked out for them.

In what follows, it may help to keep in mind that A... is Group 1, G... is Group 2, M... is Group 3, and S... is Group 4. Let there be certain magnitudes A, B, C, D, E, F have the same ratio when taken pairwise as other magnitudes equal in number, G, H, I, J, K, L, i.e., let A have the same ratio to B which G has to H, and B to C, which H has to I, and the rest similarly.  And let A, B, C, D, E, F be picked out in any ratio to other magnitudes, M, N, O, P, Q, R, with G, H, I, J, K, L in the same ratio correspondingly to other magnitudes, S, T, U, V, W, X, i.e., let G to S have the same ratio as A to M, and let H to T have the same ratio as B to N, and the rest similarly.  We have to show that all of A, B, C, D, E, F have the same ratio to all of M, N, O, P, Q, R which all of G, H, I, J, K, L have to all of S, T, U, V, W, X.

Since M has the same ratio to A which S has to G, and A to B which G has to H, and B to N which H has to T, M has the same ratio to N which S has to T (by ex aequali).  And for the same reasons, N has the same ratio to O which T has to U, and similarly for the others.  (summation step 1) But all of A, B, C, D, E, F have the same ratio to A which all of G, H, I, J, K, L have to H, while A has the same ratio to M which G has to S, and (summation step 2) M to all of M, N, O, P, Q, R has the same ratio which S has to all of S, T, U, V, W, X.  Hence, (ex aequali) it is clear that all of A, B, C, D, E, F have the same ratio to all of M, N, O, P, Q, R which all of G, H, I, J, K, L have to all of S, T, U, V, W, X.

[Corollary]
It is obvious that if A, B, C, D, E, F some magnitudes are picked out, e.g., A, B, C, D, E in a ratio to M, N, O, P, Q, but F is not picked in ratio to one of them, then if out of G, H, I, J, K, L, some, i.e., G, H, I, J, K are picked out in ratio to S, T, U, V, W, but M is not picked in ratio to one, similarly all of A, B, C, D, E, F to M, N, O, P, Q will have the same ratio which all of G, H, I, J, K, L have to , T, U, V, W.

Observations

Why the complicated form:
The theorem is for geometry, not arithmetic.  As a result there may be no ratio of A1 : B1  or any sense to speaking of the ratio (A1+ ... + An) : (B1+ ... + Bn), e.g., where the A's are planes and the B's lines.  If, however, one can pick the A's as equal and the B's as equal, one can simplify the situation immensely.  Archimedes does this in the Method.  The result is that:

A : Ci = B : Di => nA : (C1+ ... + Cn) = nB : (D1+ ... + Dn)
A simpler special case (of no particular interest):
If the ratio of A1 : B1 exists, then one can infer directly from Assumption (a):
a)  Ai : Aj = Bi : Bj
that
(A1+ ... + An) : (B1+ ... + Bn) = A1 : B1
Just take line (6) of the proof and use alternando (a : b = c : d => a : c = b : d).
Again, since we are doing geometry, this would at best be a special situation.

The nature of the proof and infinitary applications.
Archimedes only proves the theorem for finite cases.  He could extend it to infinite cases by proving an infinitary version of:

If for all i, Ai : A1 = Bi : B1, it follows that (A1 + ...) : A1 = (B1 + ...) : B1
This would quickly lead him to the problem of convergent and divergent series, a direction he might not want to go. Obviously, if the series, A1 + ..., is divergent, then the ratio, (A1 + ...) : A1 becomes meaningless. For example, in the Method, this would imply that the ratio of a plane to a line is meaningful.

Note that one might think that the problem in the infinite case of equinumerosity is dealt with by starting with quadruples, but this won't save one from paradoxes resulting from summations of other quadrupling over the same set.  One needs further geometrical rules for restricting the corresponding terms, and these must make mathematical sense--a traditional problem.

In On Conoids and Spheroids, he does not need an infinitary version, since the theorem is used only in reductio ad absurdum, i.e., props. 25-29.  Here one only needs to show that if the required relation does not hold, then there is some n such that A1, .., An, and B1, .., Bn, etc. would be in ratios (A1+ ... + An) : (C1+ ... + Cn) = (B1+ ... + Bn) : (D1+ ... + Dn) and that this is impossible.

The fact that Archimedes uses the theorem in a transfinite case in the Method 14 is itself uninteresting, since he knows nothing about Cantor's theorem.  However, he clearly thinks that it is intuitively obvious that the theorem applies in the case in question, which is an infinite case.  And this is very interesting, especially as he does not seem to have a means of proving it.