me number of triangles, similar and similarly, 
bh odd, + Seneipae® Gl te ergepe 
bE decd iW Orrtin skets Vv" thet Ay HBr ert eit 
EF beeadns seed Set BRA Gy coeds pes 79 
sgeakeiphns:: ap rt [easuRe oF THE 
‘ ‘ ‘2 i : 4 ye } d 
be Perk : +3 sty hey ree of 7 | j ; 
gon, which is at the same time 
, is called a regular polygon. 
Prop. I.” Proprem. | 
_ .» Todnseribea square inva given circle. 
7 ma te - hy nied * 4 
_ Draw: two diameters AC, BD, at right angles to each 
~ other; join their extremities A, B,C, D ; and the fis 
gure Abc shall be the inscribed square: For the an- 
gles ABC, BCD, &c. are right angles, (19. 2.), and the 
chords AB, BC, &c. are equal. f ; 
Prop. Il. Prop. on 
;  Derrinirton. Aj 
a 
To inscribe a regular hexagon and an equilateral tri- 
__ angle in a given circle, 
From any point Bin the circumference, apply BA 
and BC each equal to the radius BO; draw the dia- 
meters AD, BE, CF, and join their adjoining extre- 
CDEF, thus formed, will be 
+ 
of the angles AOB; BOC will be one third of two right 
4. Cor, 24, 1.) ; therefore COD will also be one- 
third of two right angles (2. 1.) ; therefore the angles 
AOF, FOE, EOD, vertical to these, will be each one- 
| _ third of two right angles, and the six angles at O will be 
J ae ; and hence the straight lines AB, BC, CD, DE, 
EF, FA will be all equal (15. 2.), and the hexagon 
will be equilateral. It will also be equiangular ; for 
the angles FAB, ABC, &c. stand each on two-thirds 
of the whole circumference; therefore they are all equal 
me 47; 2-) “If Pa me lines be drawn joining A, C, E, the 
vertices of the alternate angles of the hexagon, there 
_ will be formed an i ae triangle sdiiertbed in the 
- eircle, as is sufficiently evident. 
Scuotium. In the same way as we have proved that 
every equilateral hexagon is equiangular, it may be 
proved that any equilateral polygon whatever in a cir- 
_ ele is also equiangular. 
Prop. III. Pros. 
_ To describe a regular decagon in a circle, also a re- 
gular pentagon. 
__. Divide the radius AO. in.extreme and mean ratio’ at 
_ the point M; take the chord AB equal to the greater 
segment OM, and AB shall be a side of the regular de- 
__ cagon, or figure of ten sides, which may be completely 
| formed by placing straight lines, each equal to AB, 
__- round the circumference. 
Join BM ; and because, by construction, AO: OM:: 
OM: AM, and AB=OM, therefore AO: AB:: AB: 
_ AM ;,hence the triangles AOB, ABM, which have an 
|. angle OAB common to both, have the sides about that 
| angle proportionals ; therefore they are similar (20. 4.) 
And because the triangle AOB is isosceles, the triangle 
\ BM is also isosceles, and AB=BM ; but AB=MO, 
construction ; therefore BM=MO, and the triangle 
BMO is isosceles. Therefore the angle AMB, which is 
equal to the sum MOB + MBO (24, 1,), will be dou- 
Bete GEOMETRY, 
221 
MAB ; therefore Regular 
Me double AOD, and Polygons 
ene ot eke But-the, 
ly les I 
wo rig! wr one-tenth 
angles, therefore AB one tenth of the cir- 
cumference ; and the chord AB is the side of a regular 
decagon inscribed in a Circle. 
If €) pened angle gf fhe Gecneesn ibe joined by 
straight lines, there wi ormed_a re, r pen 0 
ACEGI inscribed in the cirele. hi iia 
Cor. By this and the foregoing problems a regular 
quindeeagon, or polygon of fifteen sides, may be in- 
scribed in a circle... For let AL be the side of a hexa- 
gon; then the arc ABL will be 4, or 44, of the whole 
ircumferenceé, and the arc AB .', or +}; of the circum- 
ence, therefore the difference of the two arcs will be 
2,,.0r yi. of the cireumference, and LB the chord of 
the are will be the side of a quindecagon. 
Scuotium, By, bisecting the are subtended by a side 
of any. polygon, another of double the number of sides 
may. be inscribed in a circle. Hence from a square, we 
may. inscribe polygons of 8, 16, 32, &c. sides; and 
from a hexagon others of 12, 24, &c. 
The square, the regular pentagon and hexagon, and 
such figures as can be formed from them in the manner 
we have described, were the only regular figures that th 
ancients could inscribe in a circle. A mathematician o 
our own times, Mr Gauss, has however shewn, that a 
regular polygon of 17 sides may be inscribed in a circle 
by drawing straight lines and circles only ; and tha 
the same is true of all polygons of which the number 
of sides is a prime number of the form 2*41. This 
formula includes figures of 3, 5, 17, 257, 65537, &c. 
sides; but the demonstration, even in the case of 17 
sides, has not yet been given on principles purely geo- 
metrical. See Disquisitiones Arithmetice, published at 
Brunswick, 1801 ; or a French translation, 1807. 
Prop. IV. Pros. 
Having given anyregular polygon ABCD, &e. inscri- 
bed in a circle, to describe a regular polygon of the same 
number of sides about the circle. : 
At H, the middle, of the are AB, draw the tangent Fig. 132. 
aH 4; dothe same,at the middle, of each of the other 
ares BC, CD, &c. these tangents shall form. by their 
intersections a regular circumscribed polygon a bcd, &c. 
similar to the inseribed polygon. 
Draw the radii OH, OI, and because OH is pe s 
dicular to the tangent @ b, (9. 2.) and alsoto the chord 
AB (schol. 6.2.), the tangent is parallel, to the chord 
(19..1.) In like manner it may be, shewn that all the 
other sides of the circumscribing figure are parallel to 
the sides of the inscribed figure. Draw a line from O to 
b; and because the right angled triangles OH, Ob1 
are equal (18,1), for they have a. common hypothenuse 
O 4, and the side OH=OI, therefore the e HO d= 
104, and the line OJ passes through the middle of the 
arc HI, that is, a line drawn from the centre to the in« 
tersection of any two sides of the circumscribing poly- 
gon passes. through the intersection of the sides paral- 
lel to, them of the inscribed polygon : And because the 
‘angles cbO anda 40 are respectively equal to CBO and 
ABO (8Cor. 21, 1.), the whole angle cb a= CBA ; in 
like manner it may be proved, shag baf = BAF, &c. 
therefore the angles of the circumscribing polygon are 
equal to those of the inscribed polygon. Again, be- 
