Euclid's second proof in Book VI depends on the theory of similar triangles. Let AHB be a line segment of length with and . Construct the circle with AB as diameter and let C be one of the two intersections of the perpendicular chord at H with the circle and denote the length CH as "h". Then, using Thales' theorem and, as in the proof of Pythagoras' theorem by similar triangles, triangle AHC is similar to triangle CHB (as indeed both are to triangle ACB, though we don't need that, but it is the essence of the proof of Pythagoras' theorem) so that AH:CH is as HC:HB, i.e. , from which we conclude by cross-multiplication that , and finally that formula_72. When marking the midpoint O of the line segment AB and drawing the radius OC of length , then clearly OC > CH, i.e. formula_73 (with equality if and only if ), which is the arithmetic–geometric mean inequality for two variables and, as noted above, is the basis of the Ancient Greek understanding of "Heron's method".