Fortification or Military Architecture

154.

To inscribe a regular pentagon within a triangle.

First, let a pentagon be constructed upon side BA, then let lines be drawn from point C toward E, F, and D. Then, from point G, let line GI be drawn parallel to FE, and line IL parallel to EB. This method uses the principle of homothety or "scaling": by drawing a pentagon on one side and projecting lines from the opposite vertex, one can scale the pentagon to fit perfectly within the triangle's boundaries.

155. then 156.

If one wishes to inscribe a triangle within a pentagon, with both being regular, it is only necessary to describe a circle around the pentagon; likewise, construct a triangle AKP within the circle from point A, whose two sides intersect at N and M; then ANM will be the required triangle.

157. then 158. Plate 16.

To construct a square within a regular pentagon: after having made AO perpendicular A line meeting another at a right angle (90 degrees). and extended it to meet BC at F, and then setting FI and AH each to be half of AF, one will have the necessary intersections at L and K. But in figure 158, having constructed a square upon AF, one will find the intersections M and L; for the figure HAEFG is similar In geometry, "similar" figures have the same shape and proportions but may differ in size. to MKEIL. Alternatively, in 159, having constructed a pentagon around the square on B, let lines AG and AF be drawn, which are not depicted here.

160. 161.

Conversely, if one wishes to construct a pentagon around the square in 161, then upon line AD let there be a triangle similar to FEI in figure 160; then at point B, let the angle LBC be equal to FGH.

162.

To construct a regular pentagon within a square.

Having constructed a pentagon upon side CA, and then having drawn lines FD and FB, one will find the side IH.

163.

To construct a square around a pentagon.

From points B and E, let perpendiculars be dropped to the extended base CD; one will then have HI as the side of the square.

Partitions of Figures. Plate 17. Figures 164. 165.

To divide the triangle ABC into three equal parts by lines parallel to AB.

In figure 165, let CD and DH each be one-third of BC. Then, let CE be the mean proportional A geometric middle value; if you have lengths A and B, the mean proportional X satisfies the ratio A:X = X:B. This is often used in geometry to find side lengths that relate to specific areas. between BC and CH, and let CF be equal to it. Then, between BC and CH, let CI or CK be the mean-