8. To find the locus of the vertices of all the triangles that have the same base and equal areas.

9. To find the locus of the vertices of all triangles that have the same base and the sum of the squares of their sides equal to a given square.

10. If straight lines drawn from a given point to a given line be cut in a given ratio, to find the locus of the point of

section.

11. To find the locus of the vertices of all the triangles that have the same base and the ratios of their sides equal.

12. If a straight line drawn from a given point, and terminating in the circumference of a given circle, be cut in a given ratio, the locus of the point of section is also the circumference of a given circle.

13. If one of the extremities of straight lines drawn through a given point be terminated in a given straight line, and their other extremities be determined, so that the rectangle under the segments of each line is equal to a given rectangle, the locus of these extremities will be the circumference of a circle; and if one of the extremities be terminated in the circumference of a circle, the locus of the other extremities is a straight line.

ON PORISMS.

A porism is an indeterminate problem, or one that admits of an indefinite number of solutions; but every indeterminate problem is not a porism. As a simple example of a porism, let it be required to find a point such, that all straight lines drawn from it to the circumference of a given circle shall be equal. This point is evidently the centre of the circle. Problems of loci and porisms are in many cases mutually convertible. The preceding problem becomes a problem of the latter kind when the centre is given, and it is required to find the locus of all the points that are at a given distance from it.

CDQ, they are every way equal (I. 8), therefore the angle PDCQDC. Therefore CD is perpendicular to PQ. But PQ is given, therefore the point D and the perpendicular DC are given, and consequently the point C is given. It may be easily proved by composition that C is the point required.

Any of the exercises given in the preceding books will serve as exercises in geometrical analysis.

ON PLANE LOCI.

If every point in a straight or curve line, and no other point, fulfil certain conditions, the line is said to be a locus of the point.

As a simple illustration of a locus, consider that of a point which is always equally distant from a given point. This is obviously a circle, whose radius is equal to that distance. So the locus of a point, which is always equally distant from a given straight line, is a line parallel to it, and at a distance from it equal to the given distance.

EXERCISES.

1. Find the locus of the vertices of all the triangles that have the same base and one of their sides of a given length. 2. Find the locus of a point that is at equal distances from two given points.

3. Find the locus of a point that is equally distant from two given lines, either parallel or inclined to one another.

4. To find the locus of the vertices of all triangles that have the same base and equal altitudes.

5. To find the locus of the vertices of all triangles that have the same base and one of the angles at the base equal. 6. To find the locus of the angular point opposite to the hypotenuse of all the right-angled triangles that have the same hypotenuse.

7. To find the locus of the vertices of all the triangles that have the same base and equal vertical angles.

8. To find the locus of the vertices of all the triangles that have the same base and equal areas.

9. To find the locus of the vertices of all triangles that have the same base and the sum of the squares of their sides equal to a given square.

10. If straight lines drawn from a given point to a given line be cut in a given ratio, to find the locus of the point of section.

11. To find the locus of the vertices of all the triangles that have the same base and the ratios of their sides equal.

12. If a straight line drawn from a given point, and terminating in the circumference of a given circle, be cut in a given ratio, the locus of the point of section is also the circumference of a given circle.

13. If one of the extremities of straight lines drawn through a given point be terminated in a given straight line, and their other extremities be determined, so that the rectangle under the segments of each line is equal to a given rectangle, the locus of these extremities will be the circumference of a circle; and if one of the extremities be terminated in the circumference of a circle, the locus of the other extremities is a straight line.

ON PORISMS.

A porism is an indeterminate problem, or one that admits of an indefinite number of solutions; but every indeterminate problem is not a porism. As a simple example of a porism, let it be required to find a point such, that all straight lines drawn from it to the circumference of a given circle shall be equal. This point is evidently the centre of the circle. Problems of loci and porisms are in many cases mutually convertible. The preceding problem becomes a problem of the latter kind when the centre is given, and it is required to find the locus of all the points that are at a given distance from it.

EXERCISES.

1. Two points being given, to find a third, through which any straight line being drawn, the perpendiculars upon it from the two given points shall be equal.

2. Three points being given, to find a fourth, through which any straight line being drawn, the sum of the perpendiculars upon it from two of the given points on one side of it, shall be equal to the perpendicular on it from the third point.

There are similar porisms when four, five, or any number of points are given, by which another point is to be found such that any straight line being drawn through it, the sum of the perpendiculars from the points on one side of it shall be equal to the sum of those from the points on the other side of it. The point so found is called the point of mean distance; or if the points be considered to be equal physical, or material points, it is called the centre of gravity.

3. A straight line and a circle being given, to find a point such that the rectangle under the segments of any straight line drawn through it, and limited by these, shall be equal to a given rectangle.

This problem is the 13th exercise on loci converted into a porism. Many other problems of loci may be similarly converted; and conversely.

ADDITIONAL SECOND BOOK.

The arrangement of the propositions in this book is nearly the same as in Euclid, and some of them are new. Although the notation be algebraical, no knowledge of algebra is requisite, and the reasoning is strictly geometrical. The enunciations of those propositions that are the same as in Euclid, are differently expressed to accommodate them to the notation. It is of some advantage to vary the enunciations, as the same proposition occurs in geometrical investigations in more than one form, and its identity in different forms

is not so easily recognised unless a familiarity be acquired with the various forms of expression. It may be remarked, that when the propositions of the second book are once understood, the figures employed by Euclid, in their demonstrations, readily suggest the proposition; but it will certainly be admitted that the expressions in this book are so obvious by the algebraical notation, that they suggest the propositions, with at least equal facility, and even more so, when the analogy between algebraical products and the expressions for rectangles is considered. The cause of this analogy will be observed in the tenth proposition in the book on the Quadrature of the Circle; and, in fact, it appears by it, that, when the symbols representing the sides of a rectangle denote the numbers proportional to its sides, the algebraical product and the expression for the area of the rectangle are identical.

PROPOSITION I.. THEOREM.

A line is equal to the sum or difference of its segments, and twice the mean distance is equal to the difference or sum of the segments, according as the point of section is internal or external.

Let AB be cut equally in C and unequally in D, then AB = ADDB and 2CD = ADDB, taking the upper sign for the first figure, and the lower

for the second.

A

ᄒ

E

For (fig. 1) AB = AD + DB, and A AD-DBAC+ CD-DB=CB+CD-DB=DB + CD + CD-DB 2CD.

=

And (fig. 2) AB AD-DB, and AD + DB = AC + CD + DB = CB+ CD + DB = 2CD.

If, of three lines, the excess of the second above the third is to be taken away from the first, the remainder will be the same as if the second be taken from the sum of the first and third.

Let A, B, C, represent three lines, then the excess of A above BC is equal to the excess of A+ C above B, or A-(B-C)=A+C-B.