Fortification or Military Architecture

...mean proportional In geometry, a mean proportional is a value that mediates between two others in a geometric progression (a:x = x:b). and lines KL, FG drawn parallel to BA: For as the squares of CB, CK, and CF [relate], so do the similar triangles: figure 164 is done the same way.

166.

To divide the quadrilateral ABCD into three equal parts parallel to AD.

Let CF be parallel to DB (then the imagined triangle FDA will be equal in area to the quadrilateral), therefore as EA is to AF, so the whole triangle ADE is to the quadrilateral: then let FA be divided into 3 equal parts at G and H, and a semi-circle be drawn upon EA to find the mean proportionals; and after having raised perpendiculars upon these two points where they intersect the said semi-circle, let arcs be drawn from center E as in I and K, by which the parallels to AD (such as IM and KL) will divide the figure as required: For as AE is to EH, so the squares of AE and EI, or the triangles AED to IEM; by conversing A logical step in a geometric proof where the terms of a ratio are swapped or rearranged to prove a related proportion. as EA is to AH (a third of AF), so AED is to AIMD, which will therefore be a third of the triangle ADF, or the quadrilateral ABCD; Marolois has drawn so many unnecessary lines, such as AS and AP and three circles, and yet with a great deal of flowery speech, he accomplishes nothing at all that one can understand.

167. 168. 169.

To divide the quadrangle ABCD into three equal parts by lines from angle D, as also the triangle from point D.

In figure 167, let the Diagonal AC be divided into three parts by F and E, and by drawing parallels from these to the other Diagonal DB, one will find points H and G, from which lines drawn to D will satisfy the requirement. In 168, let A3 be parallel to DB and divide C3; then from points outside the figure (such as 2), let 2F be parallel to DB; then the lines from D toward F and 1 will perform the partition. Now, for triangle 169, let AB be divided into three by points 1 and 2, from which parallels to CD will show the points through which one must draw DF and DE.

170.

To divide a pentagon through point A into 3 equal parts.

Let BG be parallel to AC, and EF parallel to AD; then divide GF into three by points 2 and 1. From these, one will draw parallels such as 2K to AC, and 1H to AD; then lines AK and AH will satisfy the problem. For the triangles AKC and A2C are equal, being on the same base AC and between the same parallel lines.

171. 172.

From point E, to divide the quadrilateral ABDC into two equal parts.

In figure 171, let ACF be a triangle equal to the quadrangle, then let AF be divided equally into two at B (which here happens to be an angle of the figure). Then draw BI parallel to EC; finally, EI will divide ABDC through the middle. The reason is that EH divides the triangle equally, therefore AEC and EHC make one half; but CIE is equal to CHE, thus AEC and CIE (which is area AEIC) will make half of the quadrangle. For figure 172, one wishes to divide it into 3 equal parts; let the triangle ACF be made as above, and the base AF di...