445 
GEOMETRY. 
lo. The drcumjereitce being divided into 24 equal parts, 
^No, 6.) tofubdivide it into 48.—Draw E#, 
e 
with a fourth pair of compares, and AB — e^ — ev with 
the firft pair. The arcs K/l4, /xN, My, vO, will be the 
forty-eighth parts of tlie circumference. 
The denionftration eafily follows from the foregoing. 
It is to be obferved, that, in order to keep the dia¬ 
gram or figures clear, tlie lines mentioned are not drawn 
upon them, nor even dotted ; but the points being marked 
on the circle, no miftake can take place. 
A circumference, then, by means of four pair of com- 
pafles, or four openings of the compafles, may b.- di¬ 
vided into 48 equal parts. With the firft pair, with an 
aperture = AB, divide the circumference into 6 parts, 
beginning at the point jw; and the arcs IF, HO, mo,fl, 
ng, will be divided in two equal parts; then, by divid¬ 
ing the circumference into fix parts beginning at the 
point y, the arcs oh, fi, nk, N G, F L, will be divided 
into two equal parts. Then, dividing the circumference 
into four parts with the third pair of compafles opening 
= Aa, and fetting out from the point jw, the remaining 
arcs, LD, ic, will be divided in two equal parts; and, 
by ftarting from the point », the arcs I C, Id, will be 
divided in like manner. Laftly, dividing the circumfe¬ 
rence again into fix equal parts with the firft pair of 
■compafles, but beginning from the laft points found by 
the third pair, all the reft of the arcs will be divided 
into two equal parts. 
The denionftration is the fame as in No. 6 . 
Befides the propofitions and demonftrations here given, 
it may not be amifs to add fome other problems with 
their folutions, omitting their demonftrations, as they 
depend upon fuch problems as have not been intro,, 
duced, and therefore could not be here given; fi nee they 
■will fliew to what extent this ufeful fpecies of geometry 
may be carried. 
II, To divide an arc, as B C, into two equal parts at G.— 
With the radius AB, 
(by which the arc 
BC to be divided w'as 
drawn,) and from the 
centres B and C,which 
are the two extremi¬ 
ties of the arCjdel’cribe 
the arcs AD and AE; 
fetoffBC=AD=AE; 
then, from the centres 
D and E, with a radius 
DC=BE, deferibe 
two arcs cutting eachpther in F. Then, with the radius 
AF and the fa,me centre^ D apd E^draw two other arcs in- 
Vql.VIJI. No..5.16. 
terfefling in G; the point G will be on the clrcumferencei 
and the arc BG will be equal to the arc GC ; confe- 
quently BC is equally divided. 
12. To divide the djlance AB in two equal parts', i.e. to 
find the point M which fiiall be 
in the middle of the right line 
AB.—Deferibe the femicircle 
BCDE ; from the centre E, 
with a radius EB, draw an 
indefinite arc PB/;; from the 13 
centre B, with a radius BA, 
draw another femicircle/)APw; 
then, from the centre P with 
the radius PB, deferibe the 
arc BM ; and from P fet off 
P?nr=:BM: M will be the point fought, midway be¬ 
tween A and B. 
13. To divide the circumference B Di/ into five equal parts .— 
Things being as in Art. j. from A as a centre fet oS' 
Aa=:N b — Ob, with the third pair of compalTes; from 
B fet off B^=:BQ; and the arc BQ will be the fifth 
part of the circumference.—The other letters placed 
round this diagram ftiew the divifions or interfedfions 
of the circumference, as demonftrated in the preceding 
articles. 
14. To find a mean and 
extreme proportional on the 
line AB.—From the cen¬ 
tre A, with the radius 
AB, deferibe the circle 
BD^<T; and in its circum¬ 
ference fet off AB=;BC 
=:CD=DE=Etf. From 
B as a centre, make BD 
z=Ba=iEa; and from A 
make A az=.E b-z=.d ^. The 
right line AB will be 
divided in mean and ex¬ 
treme proportion at the 
point b-, and we have 
BA: Ab'.: Ab ibB. 
15. The arc BCD, deferibedfrom the centre A, being given, 
to find the two points where the circumference is cut by the right 
$ N Une 
