418 A Demonstration of 
because FD : DG: : FH: GH, therefore 
Dm’: Do : : Em: Eo, and FE: EG:: 
FD:DG::FE: Ei. Therefore if in any 
line Emo, be taken two points D E such, 
that Dm : Do: : Em: Eo, and m F,oG be 
drawn perp. to Em, and through the point 
D, be drawn any line to meet mF, oG in F 
and G,and EF, EG be joined; then FE : 
6 EG::FD : DG, and FE: EG:: FE: 
Ei. Which is the sixth Proposition. 
Since Gi is bisected by the perp. Do, .-. 
Di= DG;; and because » DEH is right, .°. 
En is equal to nH, and parallel to iD ; be- 
cause iG is parallel to KH, therefore the 
lines FG, FH. are similarly divided in the 
—-- 
Cand D be assumed such that AC: CB:: AD: DB, and 
from D an indefinite perpendicular to the same diameter as 
L Dbe erected, and through C any line be drawn to cut 
the same in E, andthe circle in F and G; I say FC: 
CG:: FE: EG. 
The converse of this proposition is also true, which is. . 
this. 
If any right line as L D be drawn perpendicular to the 
diameter A B of any circle and meets the same in D, and 
if from a point in the same diameter, as C, any line be 
drawn to meet the same perpendicular in E, and the circle 
in F andG, so that FC: CG: : FE: EG; Isay that AC: 
CB:: AD; DB. 
Prop. III, Let there be a triangle A BC, whose base 
BC is bisected in D, and through the vertex Aa line A E 
