Book II. of HE, EG are equal to the square of GH. therefore the rectangle BE, EF together with the square of EG is equal to the squares of HE, EG. take away the square

C. 47. I.

H

gle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the square of EH. but BD is equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the square of EH. wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EH. Which was to be done.

Book III.

THE

ELEMENTS

OF

EUCLID.

BOOK III.

DEFINITIONS.

EQUAL circles are

I.

those of which the diameters are equal, or from the centers of which the straight lines to the circumferences are equal.

This is not a Definition but a Theorem, the truth of which is evident; for if the circles be applied to one another, so that their centers coincide, the circles must likewife coincide, fince • the straight lines from their centers are equal.'

Book III.

VI.

A fegment of a circle is the figure con

tained by a straight line and the cir-
cumference it cuts off.

VII.

" The angle of a segment is that which is contained by the straight

" line and the circumference."

VIII.

An angle in a segment is the angle con-
tained by two straight lines drawn
from any point in the circumference
of the segment, to the extremities of
the straight line which is the base of
the segment.

IX.

And an angle is said to insist or stand
upon the circumference intercepted
between the straight lines that con-
tain the angle.

The sector of a circle is the figure con-
tained by two straight lines drawn
from the center, and the circumfe-
rence between them.

ΧΙ.

Similar fegments of a circle,
are those in which the an-
gles are equal, or which
contain equal angles.

PROP. I. PROB.

O find the center of a given circle.

Let ABC be the given circle; it is required to find its center. Draw within it any straight line AB, and bisect it in D; from the point D draw DC at right angles to AB, and produce it to E, and bisect CE in F. the point F is the center of the circle ABC.

4

For if it be not, let, if possible, G be the center, and join GA, Book III.

GD, GB. then because DA is equal to DB, and DG common

another, each of the angles is a right

E

angle d. therefore the angle GDB is a right angle. but FDB is d.ro.Def.r. likewife a right angle; wherefore the angle FDB is equal to the angle GDB, the greater to the less, which is impossible. therefore Gis not the center of the circle ABC. in the fame manner it can be shewn, that no other point but F is the center; that is, F is the center of the circle ABC. Which was to be found.

COR. From this it is manifest, that if in a circle a straight line bisect another at right angles, the center of the circle is in the line which bisects the other.

I

PROP. II. THEOR.

F any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.

Let ABC be a circle, and A, B any two points in the circumference; the straight line drawn from A to B shall fall within the circle.

For if it do not, let it fall, if possible, without, as AEB; find a D the center of the circle ABC, and join AD, DB, and produce DF any straight line meeting the circumference AB, to E. then because DA is equal to DB, the angle DAB is equal to the angle DBA; and because

D

F

C

a. I. 3.

AEB

EB

с. 16. 1.

d. 19. 1.

Book III. AE a fide of the triangle DAE is produced to B, the angle DEB is greater than the angle DAE; but DAE is equal to the angle DBE, therefore the angle DEB is greater than the angle DBE. but to the greater angle the greater side is opposite d; DB is therefore greater than DE. but DB is equal to DF; wherefore DF is greater than DE, the less than the greater, which is impoffible. therefore the straight line drawn from A to B does not fall without the circle. in the fame manner, it may be demonstrated that it does not fall upon the circumference. it falls therefore within it. Wherefore if any two points, &c. Q. E. D.

I

Fa straight line drawn thro' the center of a circle, bisect a straight line in it which does not pafs thro' the center, it shall cut it at right angles. and if it cuts it at right angles, it shall bisect it.

Let ABC be a circle; and let CD a straight line drawn thro' the center bifect any straight line AB, which does not pafs thro the center, in the point F. it cuts it also at right angles.

2. 1. 3.

b. 8. 1.

Take a E the center of the circle, and join EA, EB. then because AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two fides in the one equal to two fides in the other. and the base EA is equal to the base EB; therefore the angle AFE is equal b to the angle BFE. but when a straight line standing upon another makes the adjacent angles equal to one another, each c.10.Def.1. of them is a right angle. therefore each of the angles AFE, BFE is a right angle; wherefore the straight line CD drawn thro'

C

E

B

the center bisecting another AB that does AF

not pass thro' the center, cuts the fame at

right angles.

D

But let CD cut AB at right angles; CD also bisects it, that is, AF is equal to FB.

The fame construction being made, because EA, EB from the center are equal to one another, the angle EAF is equal d to the angle EBF; and the right angle AFE is equal to the right angle BFE. therefore in the two triangles EAF, EBF there are two an