PROBLEM G.

To bisect the circumference of a circle.

Through the centre C of o AFG draw any diameter AB.

Then AB shall bisect the Oce.

For the may be taken up, inverted and applied to its former position, so that AB may fall upon its former position.

F

COR. To divide the circumference of a circle into 4, 8, 16,

32, &c. equal parts.

If through C we draw FCG ↓

to ACB,

then FCG will bisect the arcs AFB, AGB.

.. the Oce will be divided into four equal parts.

If we now bisect the s at the centre, each of the arcs AG, GB, BF, FA will be bisected;

and.. the Oce will be divided into eight equal parts;

and so on.

PROBLEM H.

To divide the circumference of a circle into three equal parts.

G

R

A

Let C be the centre of the given circle.

With any point A in the Oce as centre and distance AC describe a PCQ;

produce AC to meet the Oce in R.

Then shall the Oce be trisected in P, Q, R.

Join CP, CQ, AP, AQ.

Then ACP, ACQ are equilateral ▲s;

=

... each of the 4s ACP, ACQ is a third of two right s;

..

LS PCR, QCR are each
and PCQ = two thirds of two right 4 s;
.. LS PCR, QCR, PCQ are = one another;

two thirds of two rights,

.. the arcs PR, QR, PQ are = one another.

COR. Hence we may divide the circumference of a circle into 6, 12, 24, &c. equal parts.

If we produce PC, QC to meet the Oce in G, E,
then will the Oce be divided into six equal arcs.

If we bisect these arcs it will be divided into 12 equal arcs; and so on.

PROBLEM K.

To divide the circumference of a circle into five equal parts.

Describe an isosceles ▲ ABC having each of the ‹ s at

the base double the at the vertex.

At the centre of the given circle draw the

(II. C)

GOH= L ABC. = GH.

The Oce may be divided into five arcs each:

For fives, each L BAC, make up two rights; ... fives, each L GOH, make up four rights;

=

.. the Oce may be divided into five arcs each = GH.

COR. Hence we may divide the circumference of a circle into 10, 20, 40, &c. equal parts.

PROBLEM L.

To divide the circumference of a circle into fifteen equal parts.

Cut off GN= one-third of the Oce, and GH = one-fifth of the Oce,

bisect the arc HN in K. Then the Oce may be divided into fifteen parts each = HK.

For of such parts, as the Oce contains fifteen, GN contains five, and GH contains three.

.. HN contains two ;

... each of the parts is

=

HK.

COR. Hence the Oce of a may be divided into 30,

60, &c. equal parts.

129

SUPPLEMENT TO BOOK III.

THEOREM (a).

If in a circle two chords which do not both pass through the centre, cut one another, they do not bisect each other,

If one of them pass through the centre it is plain that it cannot be bisected by the other which does not pass through the centre.

But if neither of them pass through the centre, as PXQ, RXS, find the centre C, and join CX.

PR

Then. CX bisects PQ, .. CX is 1 toPQ; (III. 2) and CX bisects RS, .. CX is 1 to RS; (III. 2)

.. 4 CXP and CXR are rights,

and are.. one another; which is impossible.

.. If in a circle two chords &c.

C. G.

9

THEOREM (6).

Of all straight lines which can be drawn from a given point, which is not the centre, to the circumference, the greatest is that which passes through the centre. And of the others, on the same side of that through the centre, the one drawn to a point on the circumference nearer to the extremity of that line is the greater.

Let A be the given point and C the centre of the given ©. Through A draw ACB and also any other straight lines

AR, AS to the Oce.

Then shall ACB be > AR and AR > AS.