the squares of EB and BD. From these equals take away the common square of EB. Therefore the remaining rectangle AD.DC is equal (I. Ax. 3) to the square of DB. Wherefore, if from any point, &c. Q. E. D.

COR.-If from any point without a circle, there be drawn two straight lines cutting it, as A B and AC, the rectangles contained by the whole lines and the parts of them without the circle, are equal to one another; viz., the rectangle BA.AE, to the rectangle CA.AF: for each of them is equal to the square of the straight line AD, which touches the circle.

A

EF

D

B

A demonstration including both cases of this proposition may be derived from Props. IV. and XVII. of Book VI. Exercise 1.-If two circles cut each other, the straight line joining the points of intersection, if produced, bisects the straight line which touches both circles. Exercise 2.-From a given point without a circle, whose distance from the circumference is not greater than the diameter, to draw a secant which shall be bisected by the circumference.

PROP. XXXVII. THEOREM:

If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; and if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square of the straight line which meets it, that straight line

touches the circle.

Let any point D be taken without the circle ABC, and from it let two straight lines DA and DB be drawn, of which DA cuts the circle and DB meets it. If the rectangle AD.DC be equal to the square of DB, DB touches the circle.

Draw the straight line DE, touching the circle ABC, in the point E (III.17). Find F, the centre of the circle (III. 1); and join FE, FB, and FD.

Because FED is a right angle (III. 18), DE touches the circle ABC. But DA cuts the circle (Hyp.). B Therefore the rectangle A D.DC is equal (III. 36) to the square of DE. But the rectangle AD.D C, is (Hyp.) equal to the square of DB. Therefore the square of DE is equal (I. Ax. 1) to the square of DB, and DE to D B. But FE is equal to FB (I. Def.

D

E

15). Wherefore the two sides DE and EF are equal to the two sides DB and BF, each to each; and the base FD is common to the two triangles DEF and DBF. Therefore the angle DEF is equal (I. 8) to the angle DB F. But DEF is a right angle. Therefore also DBF is a right angle (Ax. 1) and BF, if produced, is a diameter. But the straight line passing through the extremity of a diameter, at right angles to it (III. 16), touches the circle. Therefore DB touches the circle A B C. Wherefore, if from a point, &c. Q. E. D.

Exercise.-If tangents to a circle be drawn through the extremities of any two diameters which intersect each other, the straight line joining the intersections of these tangents will pass through the centre of the circle.

G

The following Propositions may be added to this Book, as Exercises on different propositions contained in it. They will include necessary references also to the previous Books.

PROP. A. THEOREM.-If any chord of a circle be produced, till the part produced be equal to the radius of the circle; and, if from the outward extremity of this secant, another secant be drawn through the centre of the circle; they will intercept arcs of the circumference, such that the convex arc is one-third of the concave arc.

PROP. B. THEOREM.-If two straight lines intersect each other, within a circle, and cut the circumference, the angle at the point of their intersection is equal to half the angle at the centre standing on the sum of the opposite arcs intercepted between them; but, if they intersect each other without a circle, and either cut the circumference or touch it, the angle at the point of their intersection is equal to haif the angle at the centre, standing on the difference of the arcs intercepted between them.

PROP. C. THEOREM.-If the circumferences of a circle be cut by two straight lines, perpendicular to each other at any point, the squares of the four segments between that point and the points where they meet the circumference, are together equal to the square of the diameter.

PROP. D. PROBLEM

-To divide a given straight line into two parts, such that the square of one of them may be equal to the rectangle contained by the other and a given straight line.

PROP. E. PROBLEM.-To draw a straight line that shall touch two circles given in position, provided the one is not wholly within the other.

PROP. F. THEOREM.-If the diameter of a given circle be produced, and two points be taken on opposite sides of the centre, such that the rectangle contained by their distances from the centre is equal to the square of the radius, any circle which passes through these points bisects the circumference of the given circle. In concluding this Book, it may be remarked that Prop. XXXVI. suggests a mode of determining the diameter of the earth. For, in the figure to the first case of that proposition, if the circumference of the circle ABC represents that of the earth, AC the diameter of the earth, CD the altitude of any mountain above the level of the earth's surface, and DB the distance of the visible horizon; it is plain that if CD and DB be given in numbers, AC may be found, in numbers, from the nature of the proposition, by an easy arithmetical computation. The diameter of the earth being thus found to be nearly 8,000 miles (say exactly 7,920 miles), it may be proved by the application of this same proposition, that the distances of the visible horizon in leagues, are very nearly as the square roots of the altitudes in fathoms; that is, supposing the altitudes of the centre of the visible horizon to be 1, 4, 9, 16, 25, &c., fathoms, the distances of the visible horizon will be very nearly 1, 2, 3, 4, 5, &c. leagues. Hence, also conversely, the altitudes in fathoms are very nearly as the squares of the distances in leagues; that is, supposing the distances of the visible horizon to be 1, 2, 3, 4, 5, &c. leagues, the altitudes of its centre will be very nearly 1, 4, 9, 16, 25, &c. fathoms. This rule, of course, applies only to altitudes within the limits of the highest mountains on the surface of the earth.

BOOK IV.

DEFINITIONS.

I.

ONE rectilineal figure is said to be inscribed in another, when all the angular points of the inscribed figure are upon the sides of

the figure in which it is inscribed, each upon each.

According to this definition, it is plain that the inscribed figure must have as many angles as the figure in which it is inscribed has sides, and consequently as many sides.

II.

In like manner, one rectilineal figure is said to be described about another, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each.

According to this definition also, it is plain that the circumscribed figure must have as many sides as the figure about which it is described has angles, and consequently as many angles.

III.

A rectilineal figure is said to be inscribed in a circle, when all the angular points of the inscribed figure are upon the circumference of the circle.

IV.

figure

A rectilineal figure is said to be described about a circle, when each side of the circumscribed touches the circumference of the circle.

According to these definitions there is no limit to the number of the sides and angles of the rectilineal figure that may be inscribed in a circle, or described about it.

V.

о

In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle touches each side of the figure.

VI.

A circle is said to be described about a rectilineal figure, when the circumference of the circit passes through all the angular points of the figure about which it is described.

VII.

A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle.

The meaning of this definition is, that to place a straight line in a circle, is to draw a chord in the circle of a given length.

A rectilinear figure or polygon which has all its sides equal to one another is called equilateral; and that which has all its angles equal to one another is called equiangular.

A porygon which has all its sides and all its angles equal to one another, it called a regular polygon. Polygons receive particular names, according to the number of their sides and angles. Thus beginning with the triangle and the trapezium, for the sake of uniformity these are called the trigon, and the tetragon; but these terms are generally restricted to the equilateral triangle and the square. A polygon of five sides, is called a pentagon; of six sides, a hexagon; of seven sides, a heptagon; of eight sides, an octagon; &c. A polygon of ten sides is called a decagon; of twelve sides, a duodecagon; and of fifteen sides, a quindecagon, or more properly a pentedecagon,

In a given circle to place a straight line, equal to a given straight line which is not greater than the diameter.

Let ABC be the given circle, and D the given straight line, not greater than the diameter. It is required to place in the circle ABC a straight line equal to D.

Find the centre of the circle ABC (III. 1), and draw the diameter BC. If BC is equal to D, what is required is done; that is, in the circle ABC, a straight line BC is placed equal to D. But if BC is not equal to D, it is greater than D (Hyp.). From CB cut off CE equal to D (I. 3). From C as centre, at the distance CE, describe the circle AEF; and join CA.

Because C is the centre of the circle AEF, CA is equal (I. Def. 15) to CE. But CE is equal (Const.) to D. Therefore CA is equal (I. Ax. 1) to D. Wherefore in the circle ABC, a straight line CA is placed equal to the given straight line D, which is not greater than the diameter of the circle. Q. E. F.

Exercise. In a given circle, to place a straight line equal and parallel to a straight line given in position, and not greater than the diameter.

In a given circle to inscribe a triangle equiangular to a given triangle.

Let ABC be the given circle, and DEF the given triangle.. It is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF.

Draw the straight line G H touching the circle in the point A (III. 17). At the point A, in the straight line AH, make the angle HAC equal (I. 23) to the angle DEF. At the point A, in the straight line A G, make the angle GAB equal to the angle DFE. Join BC; the triangle ABC is the triangle required. Because GH touches the circle ABC, and

D

H

is a chord drawn from the point of contact the angle HAC is

equal (III. 32) to the angle ABC in the alternate segment of the circle. But the angle HAC is equal to (Const.) the angle DEF. Therefore also the angle ABC is equal (I. Ax. 1) to DEF. For the same reason, the angle ACB is equal to the angle DFE. Therefore the remaining angle BAC is equal (I. 32 and Ax. 1) to the remaining angle EDF. Wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Q. E. F.

Exercise. If a triangle be inscribed in one of two concentric circles, equiangular to a given triangle, it is required to inscribe the same in the other circle, so that its sides may be parallel to the sides of the former.

PROP. III. PROBLEM.

About a given circle to describe a triangle equiangular to a given

triangle.

Let ABC be the given circle, and DEF the given triangle. It is required to describe about the circle ABC a triangle equiangular to the triangle DEF.

A

K

Produce BF both ways to the points G and H. Find the centre K of the circle ABC (III. 1), and from it draw any straight line KB. At the point K in the straight line KB, make the angle BKA equal (I. 23) to the angle DEG, and the angle BKC equal to the angle OFH. Through the points A, B, and C, raw (III. 17) the straight lines LM, MN, and NL, touching the circle ABC. The triangle LMN is the triangle required.

M B

N GE

FH

Because the straight lines LM, MN, and NL touch the circle ABC in the points A, B, and C, and the straight lines KA, KB, and K C, are drawn from the centre to the points of contact, the angles at the points A, B, and C (III. 18) are right angles. Because the four angles of the quadrilateral figure AMBK are equal to four right angles (I. 32, Cor. 8), and the two angles KAM and KBM are (Const.) right angles. Therefore the two angles AKB and AMB are equal (I. Ax. 3) to two right angles. But the angles DEG and DEF are equal (I. 13) to two right angles. Therefore the two angles AKB and AMB are equal (I. Ax. 1) to the two angles DEG and DEF. But the angle AKB is equal (Const.) to the angle DEG. Therefore the remaining angle AMB is equal (I. Ax. 3) to the remaining angle DEF. In like manner it can be shown that the angle LNM is equal to the angle DFE. Therefore the remaining angle MLN is equal (I. 32 and Ax. 3) to the remaining angle EDF. Wherefore the triangle LMN is equiangular to the triangle DEF; and it is described about the circle A B C. Q. E. F.

The construction and demonstration of this propition would be as easily effected by first inscribing in the circle, a triangle equiangular to the given one, and then drawing tangents to the circle, through its three angular points. In the preceding demonstration, to prove that the tangents ML and NL must meet, it is only necessary to join AC, and apply the 9th and 12th Axioms of Book I In the same way, it can be shown that MN and ML, as well as LN and MN, must meet. The construction might be shortened by producing B K, and making the angles which the part produced makes with KA and K C, respectively