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0 47.1.

Book
Ill to the square of GH: But the squares of HE,EG are equal.c

to the square of GH: Therefore the rectangle BE, EF, to-
gether with the square of EG, is equal to the squares of
HE, EG; Take away the square of EG, which is common
to both; and the remaining rectangle BE, EF is equal to
the square of EH: But the rectangle 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.

1

THE

ELEMENTS

OF

EUCLID

BOOK III.

DEFINITIONS.

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I. EQUAL circles are those of which the diameters are equal, Book II. or from the centres of which the straight lines to the circumferences are equal.

"This is not definition, but a theorem, the truth of
which is evident; for, if the circles be applied to one an-
other, so that their centres coincide, the circles must like-
wise coincide, since the straight lines from the centres
are equal.'

II.
A straight line is said to

touch a circle, when it
meets the circle, and
being produced does not
cut it.

III.
Circles are said to touch one

another, which meet but
do not cut one another.

IV.
Straight lines are said to be equally

distant from the centre of a circle,
when the perpendiculars drawn to
them from the centre are equal.

V.
And the straight line on which the

greater perpendicular falls, is said to be further from the centre.

Book III.

VI.
A segment of a circle is the figure

contained by a straight line and
the circumference 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

contained by two straight lines
drawn from any point in the cir-
cumference 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 orstand

upon the circumference intercept-
ed between the straight lines that contain the angle.

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See N. To find the centre of a given circle.

Let ABC be the given circle; it is required to find its centre.

Draw within it any straight line AB, and bisecta it in D; b11. 1. from the point D drawb DC at right angles to AB, and pro

duce it to E, and bisect CE in F: The paint F is the centre of the circle ABC.

a 10.1.

F

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For, if it be not, let, if possible, G be the centre, and join Book III. GA, GD, GB: Then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG, are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are drawn from the centre G*: Therefore the

G angle ADG is equal to the angle GDB: But when a straight line standing upon another straight line ΑΣ

B makes the adjacent angles equal to one another, each of the angles is a right angled: Therefore the angle GDB is a right angle: d 10 Def. 1 But FDB is likewise a right angle: wherefore the angle FDB is equal to the angle GDB, the greater to the less, which is impossible: Therefore G is not the centre of the circle ABC. In the same manner it can be shown, that no other point but F is the centre; that is, F is the centre 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 centre of the circle is in the line which bisects the other.

a

PROP. II. THEOR.

IF 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

a 1. 3. centre of the circle ABC; and join

D AD, DB, and produce DF, any straight line meeting the circumference AB to E: Then because DA is equal to DB, the angle DAB 's equal

15. 1. to the angle DBA; and because AE,

* N. B. Whenever the expression.“ straight lines from the centre" or “ drawn from the centre” occurs, it is to be understood that they are drawn to the circumference.

€ 16.1.

Brox III. & side of the triangle DAE, is produced to B, the angle

DEB is greater than thre 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 • 19. 1. is opposited; 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 impossible: Therefore thestraight line drawn from A to B does not fall without the circle. In the same 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.

PROP. III. THEOR.

Ir a straight line drawn through the centre of a cirele bisect a straight line in it which does not pass through the centre, 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 through the centre, bisect any straight line AB, which does not pass through the centre, in the point F: It cuts it also

at right angles. 2 1. 3.

Takea Ethe centre 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 sides in the one equal to two sides in the other, and the basė

EA is equal to the base EB; there-
8. 1. fore the angle AFE is equal b to the

angle BFE: But when a straight
linestanding upon another makes the

adjacent angles equal to one another, <10 Def. 1. each of them is a right angle: There

fore each of the angles AFE, BFE, is
a right angle; wherefore the straight

A


line CD, drawn through the centre,
bisecting another AB that does not

D
pass through the centre, cuts the same at right angles.

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

The same construction being made, because EA, EB,

from the centre are equal to one another, the angle EAF * 5.1. is equald to the angle ÉBF: and the right angle AFE is

equal to the right angle BFE: Therefore, in the two tri

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