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21. Theorem. Let ACB be a catenary Fig. 25 .whose vertex is C, axis CD, and directrixEHj take any two equal parts FG, GH, onthe directrix, and parallel to CD, drawFM, GN, and HB, through F, G, and H,cutting the curve in M, N, and B ; throughthese points again draw the double ordin-ates MO m, NP n, and BQA. On the pro-longation of FM, GN, and HB, take MI= CM,NK=CN, and BL=CB ; join 10, KP, LQ;and produce them to meet the perpendicularsmf, ng, Ah, drawn from m, n, and A, in i, k,and l. Then will CE be a mean proportionalbetween FI and fi, between GK and gk, andbetween HL and hi ; gk will be a meanproportional between fi and hi.

For since the two triangles IMO, imo, areequi-angular, and OM =om, MI must beequal to mi. In the same manner it maybe shown that NK = w&, and BL=A£

Hence by adopting the same notation asbefore FM = fin = t, and MI = mi — CM = z;whence FI — t+z and fi — t—z, consequentlyFIx^ = (jf+ 2 :) (if— z)—t 2 '-~z z . But by art. 9.

Case X., consequently FI xfi—c z

K