BY means of this Consectary Nicomedes solves that noble Problem of finding two mean proportionals, after this way, which we will here shew from Eutocius, but drawn into a compendium, and somewhat changed as to the order. Let two given lines AB and BC (Fig. 144.) between which you are to find two mean proportionals, be joined together at right angles, and divide both into two parts in D and E, and ha∣ving compleated the rectangle ABCL, from L thro' D draw LG to BC prolonged; so that after this way GB may become = AL or BC: Having let fall a perpendicular from E cut off from C at the interval CF=AD the part EF, and having drawn FG make CH parallel to it; and lastly thro' the legs of the angle KCH draw the right line FHK, so that the part HK shall be equal to the line CF, by the preceding Consect. and also draw the right line KM from K thro' L to the con∣tinued line BA: All which being done, CK and AM will be two mean proportionals between AB and BC; which after our way we thus demonstrate: By reason of the similitude of the ▵ ▵ MAL and LCK
MA is to LC or AB as AL or BC to CK
b−eb−c−ec and moreover,
as MA to AD so GC to CK i. e. FH to HK
b−½eb−2c−ec by reason of GF and CH be∣ing parallel, by Consect. 4. Prop. 34. lib. 1. therefore since HK is = AD=½eb, FH will be = A=b, and consequent∣ly MD=FK, viz. both b+½eb, and the square of both =bb+eb+¼eebb=□EF+EK by vertue of the Pythag. Theor. Now if to these equal quantities you add the equal □□ DX and EC=¼cc, their sum, viz. □MD+□DX i. e. □ MX will be bb+ebb+¼eebb+¼cc, equal to the sum of these, viz. □EF+□EC i. e. □ CF (by the Pythag. Theor. or EX by Construct.) +□KX; whence these two things now follow: 1. That the lines MX and KX are equal. 2. If from those equal sums you take away the common quan∣tities