44 INVESTIGATION of fome 



Ok — Ok. Wherefore O^ -f- Oa + Ok" = Ob* + Od? -f 3 m X 



Ob X Ok — 3Xm+ 1 xOd X0k+ 3m 2 xOb XOk +3X 



rrrS 



m+i X Od X Ok -f- Ok . Now, let this be — Ob + Od ± 



V*. Then 3 m X Ob — 3 X m+ 1 X Od -j- 3 m z x Ob X Ok + 



2 V ! 



2,X m+ 1 xOdxOk-{-Ok zz +• q^, and ?nxOb — m+ 1 



V* 



X Od + — , when 0£ = o. But m X Ob =z m -f- 1 X Od ; 



therefore V == o. For when O k — o, O 3 is the line of 72 °, and 

 O d the fine of 36 °. When O £ is a maximum, it is the fine of 

 1 8°, Oc is r= /*, Ob is the cofine of 36% and Of — 0£ the 



verfed fine of 36 °. Wherefore, #+ i*fjfe = Ob* : 05" = ver- 

 fed fine of 36 -f fine of 18 : verfed fine of 36 . 



Let BD (PI. II- fig. 3) = BH == fide of an infcribed pentagon j 

 bifea BD in F, and draw OFG, AC, BC and DH. Then, fince the 

 angle FOB is $6°, CF is the verfed fine of 36 , OG is the fine of 

 1 8°. But fince the triangles CFB, DGO, are fimilar OG : CF = DG: 



FB = DH : DB, and OG -f CF : CF = DH + DB : DB = DH* : 



DB = DG : FB = fquare of the fine of 72 ° : fquare of the 



fine of 36 °. For when DH 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 triangle MDN to the triangle BOC. 



This demonftration, however, was unnecefTary. 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 



