"He next inscribed half a regular hexagon ABCD in a semicircle whose centre was O, and on OA, AB, BC, and CD as diameters described semicircles of which those on OA and AB are drawn in the figure [2]. Then AD [by equilateral triangles within the half-hexagon] is double any of the lines OA, AB, BC, and CD,\therefore\;square\;on\;AD = sum\;of\;sqs.\;on\;OA, AB, BC, and\;CD,\therefore\;area\;\frac{1}{2} \bigodot on\,ABCD = sum\;of\;areas\;of\;\frac{1}{2} \bigodot s\;on\;OA, AB, BC, and\;CD.Take away the common parts\therefore\;area\;trapezium\;ABCD = 3\;lune\;AEBF + \frac{1}{2} \bigodot on\;OA.If therefore the area of this latter lune be known, so is that of the semicircle described on OA as diameter."