Scholia to Theodosius, Sphaerica iii 9 (p. 148.18-19), pp. 193.20 - 194.9
(general diagram)

(diagram 1) If three magnitudes of the same kind are supposed, AB, G, DE, with AB larger than G and DE of any size, let it be required to find some magnitude smaller than AB, larger than G and commensurate with DE.

(diagram 2) Let BZ be equal to G. (diagram 3 = general diagram) Bisecting DE and bisecting the half of that and repeatedly doing this, we are left with some magnitude less than AZ. Let there be left DH and let it be less than AZ, while it is a measure of DE.

DH either measures BZ or it doesn't measure it.

(diagram 4) First let it measure it and let ZQ be equal to DH. Since DH measures ZB, and DH is equal to ZQ, therefore DH measures BQ. But it also measured DE. Therefore BQ is commensurable with DE, while being smaller than AB and larger than G.

(diagram 5, with length G changed for the case) Let DH not measure ZB, and let DH in measuring out ZB exceed it by ZQ which is less than DH (and therefore ZQ < AZ). Therefore, DH measures BQ. But it measured DE. Therefore, BQ is commensurable with DE, while being smaller than AB and larger than G.
This lemma is required by all versions of two-step exhaustion proofs, including Archimedes, On the Equilibrium of Planes 6-7; Theodosius, Sphaerica iii 9, iii 10; and Pappus, Mathematical Collection v 12 (= Pappus, Commentary on Ptolemy's Almagest vi 253.4-258.12), vi 7-9 (generalization of Theodosius, Sphaerica iii 5).