(diagram 1) Again, let AG be a balance and its middle B, and let KDTR be a trapezoid so that KD, TR are sides converging to G, and DT, KR perpendiculars to BG, and let it be suspended from the balance at E, H, and let area Z be suspended at A and let it incline equally to trapezoid DKTR, holding as it is now positioned, and let trapezoid DKTR have this ratio to area L, that which AB has to BE, and let the same trapezoid have this ratio to M, that which AB has to BH. Similarly to the previous, it will be proved that Z is larger than L and smaller than M.

In this sample proof, I repeat Archimedes argument from QP 10, with only such changes as are necessary.

(diagram 2) For I took the center of the weight of trapezoid DKTR. Let it be Q. It will be taken similarly to that previously. And I draw QI parallel to DE. (diagram 3) And so if the trapezoid is suspended from the balance at I, but is released from EH, it will maintain the same situation and will incline equally to Z for the same reasons as those previously. (diagram 4) But since the trapezoid when suspended at I inclines equally to Z when suspended at A, the trapezoid has the same ratio to Z which AB has to BI. And so it is clear that DKTR has a larger ratio to L than to Z, but a smaller to M than to Z. Thus Z is larger than L and smaller than M

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