Fortification or Military Architecture

...equal to BC and drawn AR cutting the perpendicular on 12 at S, and then BT mean proportional A geometric mean: a value $x$ such that $a:x = x:b$. In this period, it was typically found by constructing a semicircle. between QB, S, and 12.
Furthermore, let $r$ be in the middle of RP, and with A $r$ extended to M (after having made BH equal to BC), finally let W be marked from center T with the interval MH; then from center W and with the same interval, let there be the arc TV. Then VB will be the required width: it will also be 28 $1/2$ — $\sqrt{426 3/5}$ according to the hypothesis found near the figure at the end of the plate; where one must also erase a 4 that is quite poorly drawn and put a 1 in its place.

From triangle ABC, to subtract one-third by a line from point F, such as FM.

Let EC be one-third of BC (BC because the point is on that side), then AEC will be one-third of the whole; then let CA be extended, meeting line PD at O (parallel to BC at the distance of the perpendicular FH); then let triangle OCI be made equal to ACE; then FG parallel to AC, met by BC at G; let GC be cut at N, so that GN, NC, and CI are proportionals Meaning they form a geometric progression. by the preceding [Proposition] 97, plate 8: finally, with FM drawn through N, triangle MNC will be the one required. For when 3 lines are proportional—GN, NC, CI—just as the squares of GN and NC (or the similar triangles GFN and MNC), so GN is to CI (or rather the triangles GFN and OCI which are of the same height); therefore triangle GFN will have the same ratio to triangle MNC as to triangle OCI, and consequently MNC will be equal to OCI, which is one-third of the whole.

Given 3 lines A, B, and C, to find a fourth such that the rectangle of A and B is equal to the rectangle of the required line on one hand, and of the combination of that line with C on the other hand.

Let AZ be equal to half of C, and a perpendicular AK [set] above it, capable of forming the rectangle of A and B: then from center Z let the arc KE be made (add the letter E, which was forgotten), then AE will be the required line. For after having made HA equal to C, Z will then be in the middle of AH; and by the 6th [proposition] of the 2nd book of Euclid, the rectangle of HEA plus The symbol † is used here as an addition sign for areas. the square of AZ will be equal to the square of EZ or of KZ, which is equal to the squares of KA and AZ. Let us remove the common square of AZ, and there will remain that the rectangle HEA is equal to the square of AK, or rather the rectangle AB.

From point K (within BA extended), to draw KL making triangle KBL equal to rectangle ABDC.

If KA is greater than AB, let the rectangle ABCD be applied—that is, divided—over half of KB, and let BL be the height found; but in Figure 181, if KE is not greater than CB, let CD then be equal to KE, and draw KD. Marolois does it otherwise, not without ceremony The translator is likely mocking Marolois for using a needlessly complex or "fussy" method..