BOOK III.

DEFINITIONS.

1.

EQUAL circles are those of which the diameters are equal, or those from the centres of which the [radii, or] straight lines drawn 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 one to another, so that their centres coincide, the circles must likewise coincide, since the straight lines drawn from the centres are equal.”

Concentric circles are those which have the same centre.

II.

A straight line is said to touch a circle, when it meets the circumference, and being produced does not cut the circle,

that is, does not intersect the circumference.

If a straight line touch a circle in the sense thus defined, it is called a tangent (touching) to the circle; but one which cuts the circle, is called a secant (cutting) to the circle.

III.

Circles are said to touch one another when their circumferences meet in any point, but do not cut one another.

The point in the circumference of a circle, where a straight line or another circle touches it, is called the point of contact (touch).

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.

The straight lines spoken of in this definition are usually called chords (strings) because they stretch from one point of the circumference to another. If the perpendiculars drawn to them from the centre, be produced to meet the circumference, the parts of these perpendiculars between the chords and the circumference are called sagittæ (arrows).

V.

And the straight line which has the greater perpendicular drawn to it, is said to be farther from the centre.

The straight line, or chord, which has the less perpendicular drawn to it, is said te be nearer to the centre.

VI.

A segment of a circle is the figure contained by a straight line or chord, and the arc, or part of the circumference which it cuts off.

Every chord, except a diameter, divides a circle into two
unequal segments, the one greater and the other less than

a semicircle. For brevity's sake, let these segments be
called supplementary to each other.

VII.

[The angle of a segment is that which is contained by the straight line and the circumference.]

This definition appears to be an interpolation. It is not used and need not be remembered.

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.

The meaning of this definition is, that if from the extremities of the chord of a segment, two other chords be drawn to any point in the arc of the segment, the angle formed between these two chords is called the angle in the segment.

IX.

An angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle.

The meaning of this definition is that the angle in a segment is said to stand upon the arc of its supplementary segment.

X.

A sector of a circle is the figure contained by two (radii, or) straight lines drawn from the centre, and the arc, or part of

the circumference between them.

The two radii, except when they are in the same straight line, and thus form a diameter, divide a circle into two unequal sectors, the one greater and the other less than a semicircle. These may also be called supplementary sectors. Sectors receive names, sometimes indicative of the part which they form of the entire circle, as a quadrant, or the

fourth part of a circle; a sextant, or sixth part; and an octant, or eighth part.

XI.

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

AXIOMS.
I.

If a point be taken between the centre of a circle and its circumference, that point is within the circle; and if a point be taken beyond the circumference, it is without the circle.

This axiom is tacitly assumed by Euclid in this Book.

II.

If two magnitudes be doubles of two other magnitudes, each of cach, the sum of the first two is double the sum of the other two.

III.

If two magnitudes be doubles of two other magnitudes, each of each, the difference between the first two is double the difference between the other two.

These two axioms are not given by Euclid; they are necessary, however, to complete the logical demonstration of the 20th proposition of this Book. Dr. Thomson, in his edition gives Playfair's demonstration of them.

Let ABC be the given circle. It is required to find its centre. Take any two points A and B, in the circumference, and join AB. Biscct the straight line AB (I. 10) at D. From the point D, draw DC at right angles (I. 11) to AB. Let CD meet the circumference in C and E; and bisect CE in F. The point F is the centre of the circle AB C. For if it be not, let, if possible, G, a point not in CE, be the centre, and join GA, GD and G B.

B

ID

E

Because DA is equal (Const.) to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG, are equal to the two sides BD, DG, each to each. But the base GA is equal (I. Def. 15) to the base G B, because they are drawn from the centre G. Therefore the angle ADG is equal (I. 8) to the angle G D B. But when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of these angles is a right angle (I. Def. 10). Therefore the angle GDB is a right angle. But FDB is likewise (Const.) a right angle. Wherefore the angle FDB is equal (I. Ax. 1) to the angle G DB; that is, the greater equal to the less, which is impossible. Therefore G is not the centre of the circle A B C. In the same manner it can be shown that no other point out of CE is the centre. Because CE is bisected in F, any other point in CE divides it into unequal parts, and cannot be the centre. Therefore no other point but F can be the centre. Wherefore F is the centre of the circle A B C. Q. E. F.

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

The simplest mode of finding the centre of a circle, is to draw any two chords and bisect them at right angles by two straight lines. The intersection of these straight lines will be the centre, by the preceding corollary.

Exercise 1.-Given the arc of a circle, to find the centre of the arc, that is, of the circle of which it is an arc.

Exercise 2.-To describe a circle that shall pass through any three points which ar not in the same straight line.

If any two points be taken in the circumference of a circle, the straight line which joins them lies within the circle.

Let A B C be a circle, and A and B, any two points in the circumference. The straight line drawn from A to B lies within the circle.

Find D the centre of the circle ABC (III. 1), and join DA and DB. In the arc A B, take any point F; join D F, and let it meet the straight line AB in E.

C

B

Because DA is equal (I. Def. 15) to D B, the angle DAB is equal (I. 5) to the angle DBA. Because AE, a side of the triangle DA E, is produced to B, the exterior angle DEB is greater (I. 16) than the interior and opposite angle DAE. But the angle DAE was proved to be equal to the angle DBE. Therefore the angle DEB is also greater than the angle DBE. But the greater side is opposite (I. 19) to the greater angle. Therefore D B is greater than DE. But DB is equal (I. Def. 15) to DF. Therefore DF is greater than DE, and the point E lies within the circle (III. Ax. 1). In the same manner it may be proved that every other point in AB lies within the circle. Therefore the straight line AB lies within the circle. Wherefore, if any two points, &c.

Q. E. D.

This demonstration differs from Euclid's, in being direct, and not proceeding by the method of reductio ad absurdum. It is preferable too, on account of the greater simplicity of the diagram.

Corollary.-A straight line cannot cut the circumference of a circle in more points than two.

If in a circle, one straight line passing through the centre of a circle bisect another within it, which does not pass through the centre, the latter is cut at right angles: and conversely, if it be cut at right angles it is bisected.

Let ABC be a circle; and let CD, a straight line passing through the centre, bisect any straight line AB, which does not pass through the the centre, in the point F. The straight line A B is cut at right angles. Take E the centre of the circle (III. 1), and join E A and E B. Because AF is equal to FB (Hyp.), and FE common to the two triangles A FE, B FE, the two sides AF, FE, in the one are equal to the two sides BF, FE, in the other, each to each. But the base EA is equal (I. Def. 15) to the base E B. Therefore the angle AFE is equal (I. 8) to the angle BFE. But when one straight line standing upon another makes the adjacent angles equal (I. Def. 10) to one another, each of them is a right angle. Therefore each of the angles AFE, BFE, is a right angle. Wherefore the straight line

E

D

CD, passing through the centre and bisecting another A B that does not pass through the centre, cuts it at right angles.

Next, let CD cut AB at right angles. The straight line AB is bisected at F.

Next only two straight lines drawn to the circumference from F, can be equal to each other, one being on each side of AD.

At the point E in the straight line EF make the angle FEH equal to the angle FEG. Join FH. FH is the only straight line that can be drawn to the circumference equal to FG.

Because GE is equal (I. Def. 15) to EH, and EF common to the two triangles GEF, HEF; the two sides GE and EF are equal to the two sides HE and EF, each to each. But the angle GEF is equal(Const.) to the angle HEF. Therefore the base FG is equal (I. 4) to the base FH. And no other straight line but FH, can be drawn from F to the circumference equal to FG. For, if possible, let FK be equal to FG. Because FK is equal to FG, and FG to FH, therefore FK is equal (I. Ax. 1) to FH; that is, a line nearer to that which passes through the centre, is equal to one more remote; which has been proved to be impossible. Therefore, if any point be taken, &c. Q. E. D.

Corollary 1.-If two chords of a circle intersect each other and make equal angles
with the diameter passing through their point of intersection, they are equal.
Corollary 2.-If two chords of a circle intersect each other and make unequal
angles with the diameter passing through the point of their intersection; that
which makes the least angle with the diameter is the greatest.
DEFINITION.-An arc of a circle is said to be concave towards a point without it,
when all the straight lines, drawn from that point. meet the hollow part or inside
of the arc; and it is said to be convex towards a point without it, when all the
straight lines, drawn from that point, meet the round part or outside of the arc.
Straight lines drawn from any point without a circle to touch the circum-
ference will determine the two points which, being joined, will divide the cir-
cumference into the concave and convex parts.

If from any point without a circle, straight lines be drawn to the circumference; of those which fall upon the concave part of the circumference, the greatest is that which passes through the centre; and of the rest, that which is nearer to the greatest is greater than the more remote ; but of those which fall upon the convex part of the circumference, the least is that which produced passes through the centre; and of the rest, that which is nearer to the least is less than the more remote: and only two of either set of straight lines are equal, one being on each side of the diameter. Let ABC be a circle, and D any point without it. Let the straight lines DA, DE, DF, DC be drawn to the circumference. Of those which fall upon AEFC the concave part of the circumference, DA which passes through the centre, 's the greatest; of the rest, DE is greater than DF, and DF than D C. But of those which fall upon HIKG, the convex part of the circumference, DG is the least; of the rest, DK is less than DI, and DI than D H.

Take M the centre of the circle ABC (III. 1), and join ME, MF, MC, MK, MI, and MH.

Because AM is equal to ME (I. Def. 15). To each of these equals, add MD. Therefore AD is is equal (I. Ax. 2) to EM and MD. But EM and MD are greater than ED (I. 20). Therefore also AD

C

H

D

M