Lastly, let the side AC be perpendicular to BC: then BC is the straight line between the perpendicular and the acute angle at B: and it is manifest that the squares of AB, BC, are equal to the square of AC, and twice the square of BC: therefore B in every triangle, &c. Q. E. D.

*

PROP. XIV. PROB.

To describe a square that shall be equal to a given 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 * 45.1. BCDE equal to the rectilineal figure A. Then if 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, A

H

+30Det.

B

produce one of them

G

BE to F, and make +

EF equal to ED,

and bisect+ BF in G; and from the centre G, +10. 1.. at the distance GB, or GF, describe the semicircle BHF, and produce DE to H. The square described upon EH shall be equal to the given rectilineal figure A.

Join GH and 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 *5.2. square of GF: but GF is equal to GH: there- +15 Def. fore the rectangle BE, EF, together with the square of EG, is equal to the square of GH: but the squares of HE, EG are equal to the *47. 1. square of GH: therefore the rectangle BE, EF,

+1 Ax. 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 re+3 Ax. maining 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 † Const. square of EH: but BD is equal † to the rectilineal figure A; therefore the rectilineal figure A +1 Ax. 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.

67

THE

ELEMENTS OF EUCLID.

BOOK III.

DEFINITIONS.

I.

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

O

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.

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 circumference of the segment to the extremities of the straight line which is the base of the segment.

IX.

D

And an angle is said to insist or stand upon the circumference intercepted beween the straight lines that contain the angle.

X.

A sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them.

XI.

Similar segments of circles are those in which the angles are equal, or which contain equal angles.

PROP. I. PROB.

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 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 shall be the centre of the circle ABC.

For if it be not, let, if possible, G be the centre, and join 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

*

FIG

*10.1.

*11. 1.

+Const.

D

B 15 Def.

E

angle ADG is equal to the angle GDB but *8. 1. when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a* right angle ; *10 Def. therefore the angle GDB is a right angle: but 1. FDB is likewise a † right angle; wherefore the +Const. angle FDB is equal to the angle GDB, the †1 Ax. greater to the less, which is impossible; therefore G is not the centre of the circle ABC. In

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