squares of AB, BC are equal to the square of AC, and twice Book II. the square of BC, and twice the rectangle BC, CD: but because BD is divided into two parts in C, the rectangle DB, BC is equal to the rectangle BC, CD and the square of BC: and f 3. 2. the doubles of these are equal: therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC: therefore the square of AC alone is less than the squares of AB, BC by twice the rectangle DB, BC. Lastly, Let the side AC be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B; and it is manifest that the squares of AB, BC are equal s to the square of AC and twice the square of BC. Therefore, in every triangle, &c. Q. E. D.

Α

g 47. 1.

PROP. XIV. PROB.

B

TO describe a square that shall be equal to a given See N. rectilineal figure.

Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A.

Describe the rectangular parallelogram BCDE equal to the a 45. 1. rectilineal figure A. If, then, the sides of it BE, ED are equal to one another, it

is a square, and what was required is now done: but if they are not equal, produce one of them BE to F, and make EF e

qual to ED, and

H

bisect BF in G;

and from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H, and join GH; therefore, because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, the rectangle BE, EF, together with the square of EG, is equal to the square of b 5. 2. GF: but GF is equal to GH; therefore the rectangle BE, EF,

c

Book II. together with the square of EG, is equal to the square of GH; but the squares of HE, EG are equal to the square of GH: c 47.1. therefore the rectangle BE, EF, together 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.

THE

ELEMENTS OF EUCLID.

BOOK III.

DEFINITIONS.

I.

EQUAL circles are those of which the diameters are equal, or Book III.

from the centres 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 centres coincide, the circles must likewise 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 farther from the centre.

Book III.

VI.

A segment 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 contain
the angle.

X.

The sector of a circle is the figure contain-
ed by two straight lines drawn from the
centre, and the circumference between
them.

XI.

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

PROP. I. PROB.

See N.

a 10. 1.

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 bisect a it in D; b 11. 1. from the point D draw b DC at right angles to AB, and produce it to E, and bisect CE in F: the point F is the centre of the circle ABC.

For, if it be not, let, if possible, G be the centre, and join B. III. GA, GD, GB: then, because DA is equal to DB, and DG

F

C

G

c 8. 1.

common to the two triangles ADG,
BDG, the two sides AD, DG are e-
qual 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
angle ADG is equal to the angle
GDB but when a straight line stand-
ing upon another straight line makes
the adjacent angles equal to one ano-
ther, each of the angles is a right an-
gled therefore the angle GDB is a
right angle: 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, Fis
the centre of the circle ABC. Which was to be found.

B

D

E

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.

d 10. def.

1.

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 possi ble, without, as AEB; find a D the centre 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 AE, a side of the triangle

C

D

a 1. 3.

F

b 5. 1.

A

E

B

⚫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.