4 INVESTIGATION of fome 
of — OF. Wherefore Oc + Oa +08 =O02 +04 43m De 
Ob x OL — 3x ma x Od X OF + 3m’ x Ob x OF +3.X 
ati xOdx OF + OF. Now, let this be= OF +04 + 
V.. Then 3m x Ob —3xm+1X Od + 3m x Obx OL+ 
as aw > { oe Sy Vv: Cc TH. 2 
3x m+1 X¥OdxOk+ OF = + oO» and mx Ob mm+t. 
> 
x Od + , when OF = But mx Ob = m+1x Od; 
therefore V=o. For when O£= 0, O} is the fine of 72°, and 
Od the fine of 36°... When O é is a maximum, it is the fine of 
18°, Oc is =7, O48 is thecofine of 36°, and Oc —O4 the 
verfed fine: of 36°! © Wherefore,’m + 1:m= OF: 6a) wee 
fed fine of 36° + fine of 18°: verfed fine.of 36°. 
Ler BD (Pl. IL.-fig. 3) = BH = fide of an infcribed pentagon ; 
bifeé BD.in F, and draw OFC, AC, BC and DH. Then, fincé the 
angle FOB is 36°, CF is the verfed fine of 36°, OG is the fine of _ 
18°. But fince the triangles CFB, DGO, are fimilar OG : CF =DG: 
FB = DH’: DB, and OG + CF:CF=DH+DB:DB=DH ; 
DB =DG:FB = fquare of the fine of 72°: fquare of the 
fine of 36°. For when DH i is cut in extreme and’ mean ratio, 
the greater part is equal to the fide of the pentagon. 
DH is cut in extreme and mean ratio in the point L, and LH 
= BD; the triangle CDP is fimilar to the triangle DOB; and 
the timiek: MDN to the triangle BOC. 
Tus demonftration, however, was unneceflary._ For if the 
fum of the cubes of perpendiculars drawn from P to the fides of 
the) pentagon, be equal to the fum of the cubes of perpendiculars 
drawn 
