32 Mr. Davies's Symmetrical 



are equal, and that Erf, dY being sides of" a rhombus are 

 equal; and hence that dG, rfE, and d¥ are equal, and d is 

 the centre of the circle M-hich passes through E, F, G. 



Prop. VII. (fig. 2.) 



The figure G^Ej/jFZ is an equiiateial hexagon whose sides 

 are parallel. 



The first part of this theorem follows inramediately from the 

 last; the three rhombi which meet in d being equilateral, and 

 the hexagon in the proposition being composed of the exterior 

 sides of these equilateral rhombi. 



The hexagon has also its opposite sides parallel, for each 

 pair of opposite sides is respectively perpendicular to the same 

 line ; viz. one pair to one line, another to another, and the third 

 to a third — [Gk and Ym to AC, Gl and Era to BC, and F / 

 and EA" to AC) — whence these three pairs of opposite sides 

 have the stated property of parallelism. 



Prop. VIII. (fig. 2.) 



. The triangle Jclm is equal in all respects to the triangle 

 EFG, and the sides of the one are parallel respectively to the 

 sides equal to them of the other. 



For. G^ is equal and parallel to F?ff (Pr. 7.) ; and therefore 

 the line km is equal and parallel to GF. In the same manner 

 it appears that kl is equal and parallel to EF, and Im to GE. 

 Hence the two triangles have their sides equal and parallel 

 each to each. 



Prop. IX. (fig. 2.) 



Let the points of contact of the inscribed circle be O, P, Z ; 

 its centre e is the centre of the triangle klm. 



For, since GlVd is a rhombus, the line rfK is perpendi- 

 cular to GF, and dlh equal to twice d K ; but E <? is also equal 

 to twice dK. (Cor. Pr, III,). Whence dl is equal to E^. 



The lines dl and E^ are also parallel, being both perpen- 

 dicular to GF. Hence le is parallel and equal to rfE. 



By similar reasoning it can be proved that JG is equal to 

 em, and ek to d¥. But rfG = rfE = rfF (Prop. 6.); and, 

 therefore le-=.me-=ike\ 



and hence e is the centre of the circle klm. 

 Prop. X. (fig. 2.) 



Let ^E meet the circumscribing circle in I, <?F in Y, eG in 

 N, GF in K, GE in Q, and EF in R. Then, first, the di- 

 agonals rf/, dm, dk of the rhombi pass through K, R, Q. 



For the circle ABC passing through the feet of the perpen- 

 diculars bisects the sides FG, GE, EF in K, Q, R. But these 



sides 



