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to be situated in some line, and hence such line is called the locus of the point which satisfies the conditions of the problem.

If any two given points A and B (fig. Prop. 5, Book iv.) be joined by a straight line AB, and this line be bisected in D, then if a perpendicular be drawn from the point of bisection, it is manifest that a circle described with any point in the perpendicular as a centre, and a radius equal to its distance from one of the given points, will pass through the other point, and the perpendicular will be the locus of all the circles which can be described passing through the two given points.

Again, if a third point C be taken, but not in the same straight line with the other two, and this point be joined with the first point A; then the perpendicular drawn from the bisection E of this line will be the locus of the centres of all circles which pass through the first and third points A and C. But the perpendicular at the bisection of the first and second points A and B is the locus of the centres of circles which pass through these two points. Hence the intersection F of these two perpendiculars, will be the centre of a circle which passes through the three points, and is called the intersection of the two loci. Sometimes this method of solving geometrical problems may be pursued with advantage, by constructing the locus of every two points separately, which are given in the conditions of the problem. In the Geometrical Exercises which follow, only those local problems are given where the locus is either a straight line or a circle.

There is another class of Propositions called Porisms. The exact meaning of the term as applied to this class of propositions, appears to be as uncertain as the derivation of the word itself. In the original Greek of Euclid's Elements, the corollaries to the propositions are called porisms (Tropionata); but this scarcely explains the nature of porisms, as it is manifest that they are different from simple deductions from the demonstrations of propositions. Some analogy, however, we may suppose them to have to the porisms or corollaries in the Elements. Pappus (Coll. Matth. Lib. vii. pref.) informs us that Euclid wrote three books on Porisms, and gives an obscure account of them. He defines “a porism to be something between a problem and a theorem, or that in which something is proposed to be investigated.” Dr. Simson, to whom is due the merit of having restored the porisms of Euclid, gives the following definition of that class of propositions: “Porisma est propositio in qua proponitur demonstrare rem aliquam, vel plures datas esse, cui, vel quibus, ut et cuilibet ex rebus innumeris, non quidem datis, sed quæ ad ea quæ data sunt eandem habent relationem, convenire ostendendum est affectionem quandam communem in propositione descriptam.” Professor Dugald Stewart defines a porism to be “A proposition affirming the possibility of finding one or more of the conditions of an indeterminate theorem.” Professor Playfair in a paper (from which the following account is taken) on Porisms, printed in the Transactions of the Royal Society of Edinburgh, for the year 1792, defines a porism to be “A proposition affirming the possibility of finding such conditions as will render a certain problem indeterminate or capable of innumerable solutions."

It may without much difficulty be perceived that this definition represents a porism as almost the same with an indeterminate problem. There is a large class of indeterminate problems which are, in general, loci, and satisfy certain defined conditions. Every indeterminate problem containing a locus may be made to assume the form of a porism, but not the converse. Porisms are of a more general nature than indeterminate problems which involve a locus.

It is highly probable that the ancient geometers arrived at all geometrical truths in their attempts at the solution of problems. The first mathematical enquiries must have occurred in the form of questions, where something was given, and something required to be done; and by reasonings necessary to answer these questions, or to discover the relations between the things that were given, and those that were to be found, many truths were suggested which came afterwards to be the subjects of separate demonstration.

The ancient geometers appear to have undertaken the solution of problems with a scrupulous and minute attention, which would scarcely allow any of the collateral truths to escape their observation. They never considered a problem as solved till they had distinguished all its varieties, and evolved separately every different case that could occur, carefully distinguishing whatever change might arise in the construction from any change that was supposed to take place among the magnitudes which were given. This cautious method of proceeding soon led them to see that there were circumstances in which the solution of a problem would cease to be possible; and this always happened when one of the conditions of the data was inconsistent with the rest. Such instances would occur in the simplest problems; but in the analysis of more complex problems, they must have remarked that their constructions failed, for a reason directly contrary to that assigned. Instances would be found where the lines, which, by their intersection, were to determine the thing sought, instead of intersecting one another, as they did in general, or of not meeting at all, would coincide with one another entirely, and consequently leave the question unresolved. The confusion thus arising would soon be cleared up, by observing, that a problem before determined by the intersection of two lines would now become capable of an indefinite number of solutions. This was soon perceived to arise from one of the conditions of the problem involving another, or from two parts of the data becoming one, so that there was not left a sufficient number of independent conditions to confine the problem to a single solution, or any determinate number of solutions. It was not difficult afterwards to perceive, that these cases of problems formed very curious propositions, of an indeterminate nature between problems and theorems, and that they admitted of being enunciated separately. It was to such propositions so enunciated that the ancient geometers gave the name of Porisms.

Besides, it will be found, that some problems are possible within certain limits, and that certain magnitudes increase while others decrease within those limits; and after having reached a certain value the former begin to decrease, while the latter increase. This circumstance gives rise to questions of maxima and minima, or the greatest and least values which certain magnitudes may admit of in indeterminate problems.

In the following collection of problems and theorems, many will be found to be of so simple a character, (being almost obvious deductions from propositions in the Elements) as scarcely to admit of the principle of the Geometrical Analysis being applied, in their solution.

GEOMETRICAL EXERCISES ON BOOK I.

PROBLEM I.

From two given points on the same side of a straight line given in position, draw two straight lines which shall meet in that line, and make equal angles with it ; also prove, that the sum of these two lines is less than the sum of any other two lines drawn to any other point in the line.

ANALYSIS. Let A, B be the two given points, and CD the given line.

Suppose G the required point in the line, such that AG and BG being joined, the angle AGF is equal to the angle BGD.

B

A

D

E

Draw AF perpendicular to CD and meeting BG produced in E.
Then, because the angle BGD is equal to AGF, (hyp.)

and also to the vertical angle FGE, (1. 15.)

therefore the angle AGF is equal to ÈGF;
also the right angle AFG is equal to the right angle EFG,
and the side FG is common to the two triangles AFG, EFG,

therefore AG is equal to EG, and AF to FE. Hence the point E being known, the point G is determined by the intersection of CD and BE.

Synthesis. From A draw AF perpendicular to CD, and produce it to E, making FE equal to AF, and join BG cutting CD in G. Join also AG.

Then AG and BG make equal angles with CD.
For since AF is equal to FE, and FG is common to the two triangles
AGF, CGF, and the included angles AFG, EFG are equal;
therefore the base AG is equal to the base EG, and the angle AGF

to the angle EGF,
but the angle EGF is equal to the vertical angle BGD,

therefore the angle AGF is equal to the angle BGD;
that is, the straight lines AG and BG make equal angles with the
straight line CD.

Also the sum of the lines AG, GB is a minimum.

For take any other point H in CD, and join EH, HB. Then since any two sides of a triangle are greater than the third side, therefore ĚH, HB are greater than EB in the triangle EHB.

But EG is equal to AG; therefore EH, HB are greater than AG, GB. That is, AG, GB are less than any other two lines which can be drawn from A, B, to any other point H in the line CD.

PROBLEM II.

To trisect a given straight line. Analysis. Let AB be the given straight line, and suppose it divided into three equal parts in the points D, E.

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On DE describe an equilateral triangle DEF,

then DF is equal to AD, and FE to EB. On AB describe an equilateral triangle ABC,

and join AF, FB. Then because AD is equal to DF, therefore the angle DAF is equal to the angle DFA, and the two angles DAF, DFA are double of one of them DAF.

But the angle FDE is equal to the angles DAF, DFA,
and the angle FDE is equal to DAC, each being an angle of an

equilateral triangle;
therefore the angle DAC is double the angle DAF;

wherefore the angle DAC is bisected by AF. Also because the angle FÅC is equal to the angle FAD, and FAD to DFA; therefore the angle CAF is equal to the alternate angle AFD:

and consequently FD is parallel to AC. Synthesis. Upon AB describe an equilateral triangle ABC, bisect

the angles at A and B by the straight lines AF, BF, meeting in F; through F draw FD parallel to AC, and FE parallel to BC.

Then AB is trisected in the points D, E.
For since AC is parallel to FD and FA meets them,
therefore the alternate angles FAC, AFD are equal;

but the angle FAD is equal to the angle FAC,
hence the angle DAF is equal to the angle AFD,

and therefore DA is equal to DF. But the angle FDE is equal to the angle CAB, and FED to CBA; (1. 29.) and therefore the remaining angle DFE is equal to the remaining

angle ACB. Hence the three sides of the triangle DFE are equal to one another,

and DF has been shewn to be equal to DA,

therefore AD, DE, EB are equal to one another. Hence the following theorem.

If the angles at the base of an equilateral triangle be bisected by two lines which meet at a point within the triangle; the two lines drawn from this point parallel to the sides of the triangle, divide the base into three equal parts.

THEOREM I.

If two opposite sides of a parallelogram be bisected, and two lines be drawn from the points of bisection to the opposite angles, these two bines trisect the diagonal. Let ABCD be a parallelogram of which the diagonal is AC.

Let AB be bisected in E and DC in F,
also let DE, FB be joined cutting the diagonal in G, H.

Then AC is trisected in the points G, H.

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Through E draw EK parallel to AC and meeting FB in K.
Then because EB is the half of AB, and DF the half of DC;

therefore EB is equal to DF; and these equal and parallel straight lines are joined towards the same parts by DE and FB; therefore DE and FB are equal and parallel. (1. 33.)

And because AEB meets the parallels EK, AC, therefore the exterior angle BEK is equal to the interior angle EAG.

For a similar reason, the angle EBK is equal to the angle AEG. Hence in the triangles AEG, EBK, there are the two angles GAE, AEG in the one, equal to two angles KEB, EBK in the other, and one side adjacent to the equal angles in each triangle, namely AE equal to EB:

therefore AG is equal to EK, (1. 26.)
but EK is equal to GH, (1. 34.)

therefore AG is equal to GH. By a similar

process,

it

may be shewn that GH is equal to HC.
Hence AG, GH, HC are equal to one another,
and therefore AC is trisected in the points G, H.

PROBLEM III.

To bisect a triangle by a line drawn from a given point in one of the sides. (Euclidis de Divisionibus.)

Analysis. Let ABC be the given triangle, and D the given point in the side AB.

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Suppose DF the line drawn from D which bisects the triangle;

therefore the triangle DBF is half of the triangle ABC.

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