GE.EH. Therefore the rectangle A E.EC is equal (I. Ax. 1) to the rectangle B E.ED. Wherefore, if two straight lines, &c. Q. É. D.

The second case of this proposition is included in the demonstration of Prop. XIV, Book II. A demonstration including all the cases may be derived from Props IV. and XVI., of Book VI.

Corollary.-If the rectangles contained by the segments of the diagonals of any quadrilateral figure, be equal, a circle may be described about it.

PROP. XXXVI. THEOREM.

If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, is equal to the square of the tangent.

Let D be any point without the circle ABC, from which the two straight lines DA and DB are drawn, of which DA cuts the circle and DB touches it. The rectangle AD.DC is equal to the square of D B.

First, let DA pass through the centre E.

Join EB, and (III. 18) EBD is a right angle.

Because the straight line AC is bisected in E, and produced to D, the rectangle AD.DC, together with the square of EC, is equal (II. 6) to the square of ED. But CE is equal to EB. Therefore the rectangle AD.DC, together with the square of EB, is equal to the square of ED. But the square of ED is equal (I. 47) to the squares of EB and BD. Therefore the rectangle AD.DC, together with the square of EB, is equal (I. Ax. 1) to the squares of EB and BD. From these equals, take

away the common square of EB. Therefore the remaining rectangle AD.DC is equal (Ax. 3) to the square of the tangent DB.

Next let DA not pass through the centre of the circle A B C.

Find E the centre of the circle (III. 1), draw EF perpendicular to AC (I. 12) and join EB, EC, and ED.

Because the straight line EF, passing through the centre, cuts the straight line AC, which does not pass through the centre, at right angles, it bisects AC (III. 3). Therefore A F is equal to FC. Because the straight line AC is bisected in F, and produced to the rectangle AD.DC together with the square of FC, is equal (II. 6) to the square of FD. To each of these equals, add the square of FE. Therefore the rectangle A D.DC, together with the squares of CF and FE is equal (I. Ax. 2) to the squares of DF and FE. But the square of ED is

equal (I. 47) to the squares of DF and FE. Also, the square of EC is equal to the squares of CF and FE. Therefore the rectangle AD.DC, together with the square of EC is equal (I. Ax. 1) to the square of ED. But CE is equal to EB. Therefore the rectangle AD.DC, together with the square of E B, is equal to the square of ED. But the squares of EB and BD are equal (1. 47) to the square of ED. Therefore the rectangle AD.DC, together with the square of EB, is equal to

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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.

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 cir cumference is not greater than the diameter, to draw a cecant which shall be bisected by the ciraunference.

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 F E, F B, and F D.

Because FED is a right angle (III. 18), DE touches the circle AB C. 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 D E is equal (I. Ax. 1) to the square of DB, and DE to D B. But FE is equal to FB (I. Def.

D

H

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 DB F 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. 1), 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 аго.

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, suo 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 cf that proposition, if the circumference of the circle A BC represents that of the earth, A C 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.

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

C

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 circle 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 polygon 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 & 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 A B C, 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.

D

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 equa` 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 straigh line given in position, and not greater than the diameter.

PROP. II. PROBLEM.

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 E GAB equal to the angle DFE. Join BC; the triangle ABC is the triangle required. Because GH touches the circle ABC, and

D

H

F

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