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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 Cbe 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 JE of this line will be

the locus of the centers 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 J5 is the locus of the centers of circles

which pass through these two points. Hence the intersection JF of

these two perpendiculars, will be the center 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 pur-

sued 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.

Whenever the quaesitum is a point, the problem on being rendered

indeterminate, becomes a locus, whether the deficient datum be of the

essential or of the accidental kind. When the quaesitum is a straight

line or a circle, (which were the only two loci admitted into the ancient

Elementary Geometry) the problem may admit of an accidentally in-

determinate case ; but will not invariably or even very frequently do so.

This will be the case, when the line or circle shall be so far arbitrary

in its position, as depends upon the deficiency of a single condition to

fix it perfectly ; â€” that is, (for instance) one point in the line, or two

points in the circle, may be determined from the given conditions, but

the remaining one is indeterminate from the accidental relations among

the data of the problem.

Determinate Problems become indeterminate by the merging of

some one datum in the results of the remaining ones. This may arise

in three diff'erent ways ; first, from the coincidence of two points ;

secondly, from that of two straight lines; and thirdly, from that

of two circles. These, further, are the only three ways in which this

accidental coincidence ol data can produce this indeterminateness ; that

is, in other words, convert the problem into a Porism.

In the original Greek of Euclid's Elements, the corollaries to the

propositions are called porisms (â– Tro^Kr/.iaTa) ; but this scarcely explains

the nature of porisms, as it is manifest that they are diff'erent 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. JPappus (Coll. Math. Lib. Vll. pref.) in-

forms us that Euclid wrote three books on Porisms. 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 fol-

lowing 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

quse ad ea quae data sunt eandem habent relationem, convenire osten-

ANCIENT GEOMETRICAL ANALYSIS. 67

dendum est affectionem quandam communem in propositione descrip-

tam." That is, " A Porism is a proposition in which it is proposed to

demonstrate that some one thing, or more things than one, are given, to

which, as also to each of innumerable other things, not given indeed,

but which have the same relation to those which are given, it is to be

shewn that there belongs some common afiection described in the

proposition." Professor Dugald Stewart defines a porism to be " A

proposition affirming the possibility of finding one or more of the con-

ditions 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 find-

ing such conditions as will render a certain problem indeterminate or

caj3able of innumerable solutions."

It may without much difficulty be perceived that this definition

represents a porism as almost the same as 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.

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 proceed-

ing 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 w^hen 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 ol an indefinite number of solutions.

This was soon perceived to arise from one of the conditions of the pro-

blem 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 indeter-

minate nature between problems and theorems, and that they ad-

mitted 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

68 ANCIENT GEOMETRICAL ANALYSIS.

certain limits, and that certain magnitudes increase while others de-

crease within those limits; and after having reached a certain value,

the former begin to decrease, while the latter increase. This circum-

stance gives rise to questions of maxima and minima, or the greatest

and least values which certain magnitudes may admit of in indeter-

minate problems.

In the following collection of problems and theorems, most 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 prin-

ciple of the Geometrical Analysis being applied, in their solution.

It must however be recollected that a clear and exact knowledge

of the first principles of Geometry must necessarily precede any in-

telligent application of them. Indistinctness or defectiveness of un-

derstanding with respect to these, will be a perpetual source of error

and confusion. The learner is therefore recommended to understand

the principles of the Science, and their connexion, fully, before he

attempt any applications of them. The following directions may

assist tiim in his proceedings.

ANALYSIS OF THEOREMS.

1. Assume that the Theorem is true.

2. Proceed to examine any consequences that result from this

admission, by the aid of other truths respecting the diagram, which

have been already proved.

3. Examine whether any of these consequences are already known

to be true, or to he false.

4. If any one of them be false, we have arrived at a reductio ad ab-

surdum, which proves that the theorem itself is false, as in Euc. I. 25.

5. If none of the consequences so deduced be known to be either

true or false, proceed to deduce other consequences from all or any of

these, as in (2).

6. Examine these results, and proceed as in (3) and (4) ; and if

still without any conclusive indications of the truth or falsehood of

the alleged theorem, proceed still further, until such are obtained.

ANALYSIS OF PROBLEMS.

1. In general, any given problem will be found to depend on

several problems and theorems, and these ultimately on some problem

or theorem in Euclid.

2. Describe the diagram as directed in the enunciation, and sup-

pose the solution of the problem effected.

3. Examine the relations of the lines, angles, triangles, &c. in

the diagram, and find the dependence of the assumed solution on some

theorem or problem in the Elements.

4. If such cannot be found, draw other lines parallel or perpen-

dicular as the case may require, join given points, or points assumed

in the solution, and describe circles if need be : and then proceed to

trace the dependence of the assumed solution on some theorem or

problem in Euclid.

5. Let not the first unsuccessful attempts at the solution of a

Problem be considered as of no value ; such attempts have been found

to lead to the discovery of other theorems and problems.

GEOMETRICAL EXERCISES ON BOOK I.

k PROPOSITION I. PROBLEM.

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 i>, E,

On DB describe an equilateral triangle DEF,

then DFis equal to AD, and FE to EB.

On AB describe an equilateral triangle ABC,

SiTi^]oii\AF,FB.

Then because ADh equal to DF,

therefore the angle AFD is equal to the angle DAF,

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 DA C, each being an angle of an

equilateral triangle ;

therefore the angle DA C is double the angle DAF',

wherefore the angle DA C is bisected by AF.

Also because the angle FA C is equal to the angle FAD,

and the angle FAD to DFA ;

therefore the angle CAFi% equal to the alternate angle AFD:

and consequently FD is parallel to A C.

Synthesis. Upon AB describe an equilateral triangle ABC,

bisect the angles at A and B by the straight lines AF, BF, meeting in F;

thi'ough i^draw FD parallel to AC, and FE parallel to BC.

Then AB is trisected in the points D, E.

For since ^ C is parallel to FD and FA meets them,

therefore the alternate angles FA C, AFD are equal ;

but the angle FAD is equal to the angle FA C,

hence the angle DAF is equal to the angle AFD,

and therefore DF is equal to DA .

But the angle FDE is equal to the angle CAB,

and FED to CBA ; (i. 29.)

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 Z>i^ 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.

70 GEOMETRICAL EXERCISES

Note. There is another method whereby a line may be divided

into three equal parts : â€” by drawing from one extremity of the given

line, another making an acute angle with It, and taking three equal

distances from the extremity, then joining the extremities, and through

the other two points of division, drawing lines parallel to this line

through the other two points of division, and to the given line ; the

three triangles thus formed are equal in all respects. This may be

extended for any number of parts, and is a particular case of Euc. VI. 10.

PROPOSITION II. THEOREM.

If two opposite sides of a parallelogram he bisected^ and two lines he drawn

from the points of bisection to the opposite angles ^ these two lines trisect

the diagonal.

Let ABCD be a parallelogram of which the diagonal is ^C.

Let AB be bisected in E, and DCm F,

also let DE, FB be joined cutting the diagonal in G, II.

Then ^ C is trisected in the points G^ H.

\

Through E draw EK parallel to ^ C 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. (l. 33.)

And because AEB meets the parallels EK, A C,

therefore the exterior angle BEK is equal to the interior angle EA G.

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 the 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, (l. 26.)

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

By a similar process, it may be shewn that GH is equal to HC.

Hence A G, GH, HC are equal to one another,

and therefore ^ C is trisected in the points G, H.

It may also be proved that BE is trisected in H and K.

PROPOSITION III. PROBLEM.

Draw through a given poiyit, between two straight lines not parallel, a

straight line which shall be bisected in that point.

Analysis. Let BC, BD be the two lines meeting in B, and let A

be the given point between them.

ON BOOK I. 71

Suppose the line EAF di-awn through A, so that EA is equal to AF\

D

I

through A draw AG parallel to BC, and G^^ parallel to EF.

Then A GHE is a parallelogram, wherefore AE is equal to GH,

but EA is equal to AFhy hypothesis ; therefore GSis equal to AF.

Hence in the triangles BHG, GAF,

the angles HBG, A GF are equal, as also BGH, GFA, (l. 29.)

also the side GHis equal to AF;

whence the other parts of the triangles are equal, (l. 26.)

therefore BG is equal to GF.

Synthesis. Through the given point A, draw AG parallel to -BC;

on GI), take GF equal to 6^^ ;

then i^ is a second point in the required line :

join the points F, A, and produce FA to meet BCin E;

then the line FE is bisected in the point A ;

draw GH parallel to AE.

Then in the triangles BGII, GFA, the side BG is equal to GF,

and the angles GBH, BGH sue respectively equal to FGA, GFA;

wherefore GH is equal to AF, (i. 26.)

but GHis equal to AE, (i. 34.)

therefore AE is equal to AFj or EF is bisected in A.

PEOPOSITION IV. PROBLEM.

From two given poirds oti the same side of a straight line given in posi-

tion, draw two straight lines which shall meet in that line, and make equal

angles with it ; also prove, that the sum of these tico 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 AGCis equal to the angle BGD.

E

Draw -4 jP 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, (l. 15.)

therefore the angle A GF is equal to the angle EGF;

72 GEOMETRICAL EXERCISES

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 ^ G^ is equal to FG, and ^i^to FF.

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 BE cutting CD in G.

Join also A G.

Then AG and BG make equal angles with CD.

For since ^i^ is equal to FE, and FG is common to the two

triangles A GF, EGF, and the included angles AFG, EFG are equal ;

therefore the base AG \^ 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 A G, GB is a minimum.

For take any other point ZTin CD, and join EH, MB, AFT.

Then since any two sides of a triangle are greater than the third side,

therefore EH, HB are greater than EB in the triangle EHB.

But EG is equal to AG, and EH to AH-,

therefore AH, HB are greater than AG, GB.

That is, A G, GB are less than any other two lines which can be

drawn from A, B, to any other point Hin the line CD.

By means of this Proposition may be found the shortest path from

one given point to another, subject to the condition, that it shall

meet two given lines.

PROPOSITION V. PROBLEM.

Giveyi one angle, a side opposite to it, ajid the sum of the other two sides,

construct the triaiigle.

Analysis. Suppose B AC the triangle required, having BC equal

to the given side, ^^4 C equal to the given angle opposite to BC, also

BD equal to the sum of the other two sides.

D

Join DC.

Then since the two sides BA, AC are equal to BD, by taking BA

from these equals, the remainder ^Cis equal to the remainder AD.

Hence the triangle A CD is isosceles, and therefore the angle ADC

is equal to the angle A CD.

But the exterior angle BAC of the triangle ADC is equal to the

two interior and opposite angles ACD and ADC:

Wherefore the angle B AC is double the angle BDC, and BDC is

the half of the angle BA C.

Hence the sjnthesis.

ON BOOK I. iO

At the point D in J^D, make the angle BDC equal to half the

given angle,

and from B the other extremity of BB, draw BC equal to the

given side, and meeting DCin C,

at C in CD make the angle DCA equal to the angle CD A, so

that CA may meet BD in the points.

Then the triangle ABC shall have the required conditions.

PROPOSITION VI. PROBLEM.

To bisett a triangle by a line drawn from a given point in, one of the sides.

Anal3^sis. Let ABC he the given triangle, and D the given point

in the side AB.

A

Suppose DFthe line di-aAvn from D which bisects the triangle;

therefore the triangle DBF is half of the triangle ABC.

Bisect BC in B, and join AB, DJE, AF,

then the triangle ABB is half of the triangle ABC:

hence the triangle ABB is equal to the triangle DBF-,

take away from these equals the triangle DBF,

therefore the remainder ADF is equal to the remainder DBF.

But ADF, DBF are equal ti'iangies upon the same base DF, and

on the same side of it,

they are therefore between the same parallels, (I. 39.)

that is, AF is parallel to DF,

therefore the point F is determined.

Synthesis. Bisect the base BC in F, join DF,

from A, di-aw ^i^ parallel to DF, and join DF.

Then because DF is parallel to AF,

therefore the triangle ADF is equal to the triangle DFF; (i. 37.)

to each of these equals, add the triangle BDF,

therefore the whole triangle ABF is equal to the whole DBF,

but ABF is half of the whole triangle ABC-,

therefore DBF is also half of the triangle ABC,

PROPOSITION VII. THEOREM.

If from a point without a parallelogram lines be draioii to the extremities

of two adjacent sides, and of the diagonal ichich they include ; of the tri-

angles thus formed, that, whose base is the diagonal, is equal to the sum of

the other two.

Let ABCD be a parallelogram of which AC is one of the diagonals,

and let P be any point without it: and let AF, PC, BP, PD be

joined.

Then the triangles APD, APB are together equivalent to the tri-

angle A PC.

E

74 GEOMETRICAL EXERCISES

Draw PGE parallel to AD or BC, and me*eting ^ J? in (r, and DC

in Â£; and join DG, GC.

Then the triangles CBP, CBG are equal: (l. 37.)

and taking the common part CBH from each,

the remainders PHB, CHG are equal.

Again, the triangles DAP, DA G are equal ; (l. 37.)

also the triangles DAG, AGO are equal, being on the same base

AG, and between the same parallels AG, DC:

therefore the triangle DAP is equal to the triangle AGO:

but the triangle PHB is equal to the triangle CHG,

wherefore the triangles PUB, DAP are equal to AGC, CHG, or

A CH, add to these equals the triangle APH,

therefore the ti-iangles APH, PHB, DAP are equal to APH, ACH,

that is, the triangles APB, DAP are together equal to the triangle

PAC.

If the point P be within the parallelogram, then the difference of

the triangles APB, DAP may be proved to be equal to the triangle

PAC.

8. Describe an isosceles triangle upon a given base and having

each of the sides double of the base, without using any i^roposition of

the Elements subsequent to the first three. If the base and sides be

given, what condition must be fulfilled with regard to the magnitude

of each of the equal sides in order that an isosceles triangle may be

constructed ?

9. In the fig. Euc. I. 5. If FC and BG meet in H, then prove

that ^JJ bisects the angle BA C.

10. In the fig. Euc. i. 5. If the angle FBG be equal to the angle

ABC, and BG, CF, intersect in O; the angle BOF is equal to twice

the angle BA C.

11.' From the extremities of the base of an isosceles triangle straight

lines are di-awn perpendicular to the sides, the angles made by them

with the base are each equal to half the vertical angle.

12. A line drawn bisecting the angle contained by the two equal

sides of an isosceles triangle, bisects the third side at right angles.

13. If a straight line drawn bisecting the vertical angle of a tri-

angle also bisect the base, the triangle is isosceles.

ON BOOK 1. 75

14. Given two points one on eadi side of a given straight line ;

. find a point in the line such that the angle contained by two lines

drawn to the given points may be bisected by the given line.

15. In the fig. Euc. I. 5, let F and G be the points in the sides

AB and reproduced, and let lines FII and 6r^be drawn perpen-

dicular and equal to FC and GB respectively : also if BH, CK, or

these lines produced meet in O ; prove that BH is equal to CK, and

BO to CO.

16. From every point of a given straight line, the straight linej;

drawn to each of two given points on opposite sides of the line are

equal : prove that the line joining the given points will cut the given

line at right angles.

17. If r be the vertex of an isosceles triangle ABC, and BA be

produced so that AD is equal to BA, and DC be drawn; shew thai

B CD is a right angle.

18. The straight line EDF, drawn at right angles to J5Cthe base

of an isosceles triangle ABC, cuts the side AB in D, and CA pro-

duced in E; shew that AED is an isosceles triangle.

19. In the fig. Euc. I. 1, if AB be produced both ways to meet

the circles in D and E, and from C, CD and CE be drawn ; the figure

CDE is an isosceles triangle having each of the angles at the base,

equal to one fourth of the angle at the vertex of the triangle.

20. From a given point, draw two straight lines making equal

angles with two given straight lines intersecting one another.

21. From a given point to draw a straight line to a given straight

line, that shall be bisected by another given straight line.

22. Place a straight line of given length between two given

straight lines which meet, so that it shall be equally inclined to each

of them.

23. To determine that point in a straight line from which the

straight lines di^awn to two other given points shall be equal, pro-

vided the line joining the two given points is not perpendicular to the

given line.

24. In a given straight line to find a point equally distant from

two given straight lines. In what case is this impossible ?

25. If a line intercepted between the extremity of the base of an

isosceles triangle, and the opposite side (produced if necessary) be

equal to a side of the triangle, the angle formed by this line and the

base produced, is equal to three times either of the equal angles of the

tiiangle.

26. In the base J5C of an isosceles triangle ABC, take a point D,

and in CA take CE equal to CD, let ED produced meet reproduced

in F, then '3.AEF= 2 right angles + AFE, or = 4 right angles + AFE.

27. If from the base to the opposite sides of an isosceles triangle,

circles which can be described passing through the two given points.

Again, if a third point Cbe 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 JE of this line will be

the locus of the centers 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 J5 is the locus of the centers of circles

which pass through these two points. Hence the intersection JF of

these two perpendiculars, will be the center 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 pur-

sued 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.

Whenever the quaesitum is a point, the problem on being rendered

indeterminate, becomes a locus, whether the deficient datum be of the

essential or of the accidental kind. When the quaesitum is a straight

line or a circle, (which were the only two loci admitted into the ancient

Elementary Geometry) the problem may admit of an accidentally in-

determinate case ; but will not invariably or even very frequently do so.

This will be the case, when the line or circle shall be so far arbitrary

in its position, as depends upon the deficiency of a single condition to

fix it perfectly ; â€” that is, (for instance) one point in the line, or two

points in the circle, may be determined from the given conditions, but

the remaining one is indeterminate from the accidental relations among

the data of the problem.

Determinate Problems become indeterminate by the merging of

some one datum in the results of the remaining ones. This may arise

in three diff'erent ways ; first, from the coincidence of two points ;

secondly, from that of two straight lines; and thirdly, from that

of two circles. These, further, are the only three ways in which this

accidental coincidence ol data can produce this indeterminateness ; that

is, in other words, convert the problem into a Porism.

In the original Greek of Euclid's Elements, the corollaries to the

propositions are called porisms (â– Tro^Kr/.iaTa) ; but this scarcely explains

the nature of porisms, as it is manifest that they are diff'erent 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. JPappus (Coll. Math. Lib. Vll. pref.) in-

forms us that Euclid wrote three books on Porisms. 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 fol-

lowing 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

quse ad ea quae data sunt eandem habent relationem, convenire osten-

ANCIENT GEOMETRICAL ANALYSIS. 67

dendum est affectionem quandam communem in propositione descrip-

tam." That is, " A Porism is a proposition in which it is proposed to

demonstrate that some one thing, or more things than one, are given, to

which, as also to each of innumerable other things, not given indeed,

but which have the same relation to those which are given, it is to be

shewn that there belongs some common afiection described in the

proposition." Professor Dugald Stewart defines a porism to be " A

proposition affirming the possibility of finding one or more of the con-

ditions 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 find-

ing such conditions as will render a certain problem indeterminate or

caj3able of innumerable solutions."

It may without much difficulty be perceived that this definition

represents a porism as almost the same as 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.

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 proceed-

ing 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 w^hen 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 ol an indefinite number of solutions.

This was soon perceived to arise from one of the conditions of the pro-

blem 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 indeter-

minate nature between problems and theorems, and that they ad-

mitted 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

68 ANCIENT GEOMETRICAL ANALYSIS.

certain limits, and that certain magnitudes increase while others de-

crease within those limits; and after having reached a certain value,

the former begin to decrease, while the latter increase. This circum-

stance gives rise to questions of maxima and minima, or the greatest

and least values which certain magnitudes may admit of in indeter-

minate problems.

In the following collection of problems and theorems, most 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 prin-

ciple of the Geometrical Analysis being applied, in their solution.

It must however be recollected that a clear and exact knowledge

of the first principles of Geometry must necessarily precede any in-

telligent application of them. Indistinctness or defectiveness of un-

derstanding with respect to these, will be a perpetual source of error

and confusion. The learner is therefore recommended to understand

the principles of the Science, and their connexion, fully, before he

attempt any applications of them. The following directions may

assist tiim in his proceedings.

ANALYSIS OF THEOREMS.

1. Assume that the Theorem is true.

2. Proceed to examine any consequences that result from this

admission, by the aid of other truths respecting the diagram, which

have been already proved.

3. Examine whether any of these consequences are already known

to be true, or to he false.

4. If any one of them be false, we have arrived at a reductio ad ab-

surdum, which proves that the theorem itself is false, as in Euc. I. 25.

5. If none of the consequences so deduced be known to be either

true or false, proceed to deduce other consequences from all or any of

these, as in (2).

6. Examine these results, and proceed as in (3) and (4) ; and if

still without any conclusive indications of the truth or falsehood of

the alleged theorem, proceed still further, until such are obtained.

ANALYSIS OF PROBLEMS.

1. In general, any given problem will be found to depend on

several problems and theorems, and these ultimately on some problem

or theorem in Euclid.

2. Describe the diagram as directed in the enunciation, and sup-

pose the solution of the problem effected.

3. Examine the relations of the lines, angles, triangles, &c. in

the diagram, and find the dependence of the assumed solution on some

theorem or problem in the Elements.

4. If such cannot be found, draw other lines parallel or perpen-

dicular as the case may require, join given points, or points assumed

in the solution, and describe circles if need be : and then proceed to

trace the dependence of the assumed solution on some theorem or

problem in Euclid.

5. Let not the first unsuccessful attempts at the solution of a

Problem be considered as of no value ; such attempts have been found

to lead to the discovery of other theorems and problems.

GEOMETRICAL EXERCISES ON BOOK I.

k PROPOSITION I. PROBLEM.

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 i>, E,

On DB describe an equilateral triangle DEF,

then DFis equal to AD, and FE to EB.

On AB describe an equilateral triangle ABC,

SiTi^]oii\AF,FB.

Then because ADh equal to DF,

therefore the angle AFD is equal to the angle DAF,

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 DA C, each being an angle of an

equilateral triangle ;

therefore the angle DA C is double the angle DAF',

wherefore the angle DA C is bisected by AF.

Also because the angle FA C is equal to the angle FAD,

and the angle FAD to DFA ;

therefore the angle CAFi% equal to the alternate angle AFD:

and consequently FD is parallel to A C.

Synthesis. Upon AB describe an equilateral triangle ABC,

bisect the angles at A and B by the straight lines AF, BF, meeting in F;

thi'ough i^draw FD parallel to AC, and FE parallel to BC.

Then AB is trisected in the points D, E.

For since ^ C is parallel to FD and FA meets them,

therefore the alternate angles FA C, AFD are equal ;

but the angle FAD is equal to the angle FA C,

hence the angle DAF is equal to the angle AFD,

and therefore DF is equal to DA .

But the angle FDE is equal to the angle CAB,

and FED to CBA ; (i. 29.)

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 Z>i^ 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.

70 GEOMETRICAL EXERCISES

Note. There is another method whereby a line may be divided

into three equal parts : â€” by drawing from one extremity of the given

line, another making an acute angle with It, and taking three equal

distances from the extremity, then joining the extremities, and through

the other two points of division, drawing lines parallel to this line

through the other two points of division, and to the given line ; the

three triangles thus formed are equal in all respects. This may be

extended for any number of parts, and is a particular case of Euc. VI. 10.

PROPOSITION II. THEOREM.

If two opposite sides of a parallelogram he bisected^ and two lines he drawn

from the points of bisection to the opposite angles ^ these two lines trisect

the diagonal.

Let ABCD be a parallelogram of which the diagonal is ^C.

Let AB be bisected in E, and DCm F,

also let DE, FB be joined cutting the diagonal in G, II.

Then ^ C is trisected in the points G^ H.

\

Through E draw EK parallel to ^ C 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. (l. 33.)

And because AEB meets the parallels EK, A C,

therefore the exterior angle BEK is equal to the interior angle EA G.

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 the 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, (l. 26.)

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

By a similar process, it may be shewn that GH is equal to HC.

Hence A G, GH, HC are equal to one another,

and therefore ^ C is trisected in the points G, H.

It may also be proved that BE is trisected in H and K.

PROPOSITION III. PROBLEM.

Draw through a given poiyit, between two straight lines not parallel, a

straight line which shall be bisected in that point.

Analysis. Let BC, BD be the two lines meeting in B, and let A

be the given point between them.

ON BOOK I. 71

Suppose the line EAF di-awn through A, so that EA is equal to AF\

D

I

through A draw AG parallel to BC, and G^^ parallel to EF.

Then A GHE is a parallelogram, wherefore AE is equal to GH,

but EA is equal to AFhy hypothesis ; therefore GSis equal to AF.

Hence in the triangles BHG, GAF,

the angles HBG, A GF are equal, as also BGH, GFA, (l. 29.)

also the side GHis equal to AF;

whence the other parts of the triangles are equal, (l. 26.)

therefore BG is equal to GF.

Synthesis. Through the given point A, draw AG parallel to -BC;

on GI), take GF equal to 6^^ ;

then i^ is a second point in the required line :

join the points F, A, and produce FA to meet BCin E;

then the line FE is bisected in the point A ;

draw GH parallel to AE.

Then in the triangles BGII, GFA, the side BG is equal to GF,

and the angles GBH, BGH sue respectively equal to FGA, GFA;

wherefore GH is equal to AF, (i. 26.)

but GHis equal to AE, (i. 34.)

therefore AE is equal to AFj or EF is bisected in A.

PEOPOSITION IV. PROBLEM.

From two given poirds oti the same side of a straight line given in posi-

tion, draw two straight lines which shall meet in that line, and make equal

angles with it ; also prove, that the sum of these tico 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 AGCis equal to the angle BGD.

E

Draw -4 jP 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, (l. 15.)

therefore the angle A GF is equal to the angle EGF;

72 GEOMETRICAL EXERCISES

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 ^ G^ is equal to FG, and ^i^to FF.

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 BE cutting CD in G.

Join also A G.

Then AG and BG make equal angles with CD.

For since ^i^ is equal to FE, and FG is common to the two

triangles A GF, EGF, and the included angles AFG, EFG are equal ;

therefore the base AG \^ 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 A G, GB is a minimum.

For take any other point ZTin CD, and join EH, MB, AFT.

Then since any two sides of a triangle are greater than the third side,

therefore EH, HB are greater than EB in the triangle EHB.

But EG is equal to AG, and EH to AH-,

therefore AH, HB are greater than AG, GB.

That is, A G, GB are less than any other two lines which can be

drawn from A, B, to any other point Hin the line CD.

By means of this Proposition may be found the shortest path from

one given point to another, subject to the condition, that it shall

meet two given lines.

PROPOSITION V. PROBLEM.

Giveyi one angle, a side opposite to it, ajid the sum of the other two sides,

construct the triaiigle.

Analysis. Suppose B AC the triangle required, having BC equal

to the given side, ^^4 C equal to the given angle opposite to BC, also

BD equal to the sum of the other two sides.

D

Join DC.

Then since the two sides BA, AC are equal to BD, by taking BA

from these equals, the remainder ^Cis equal to the remainder AD.

Hence the triangle A CD is isosceles, and therefore the angle ADC

is equal to the angle A CD.

But the exterior angle BAC of the triangle ADC is equal to the

two interior and opposite angles ACD and ADC:

Wherefore the angle B AC is double the angle BDC, and BDC is

the half of the angle BA C.

Hence the sjnthesis.

ON BOOK I. iO

At the point D in J^D, make the angle BDC equal to half the

given angle,

and from B the other extremity of BB, draw BC equal to the

given side, and meeting DCin C,

at C in CD make the angle DCA equal to the angle CD A, so

that CA may meet BD in the points.

Then the triangle ABC shall have the required conditions.

PROPOSITION VI. PROBLEM.

To bisett a triangle by a line drawn from a given point in, one of the sides.

Anal3^sis. Let ABC he the given triangle, and D the given point

in the side AB.

A

Suppose DFthe line di-aAvn from D which bisects the triangle;

therefore the triangle DBF is half of the triangle ABC.

Bisect BC in B, and join AB, DJE, AF,

then the triangle ABB is half of the triangle ABC:

hence the triangle ABB is equal to the triangle DBF-,

take away from these equals the triangle DBF,

therefore the remainder ADF is equal to the remainder DBF.

But ADF, DBF are equal ti'iangies upon the same base DF, and

on the same side of it,

they are therefore between the same parallels, (I. 39.)

that is, AF is parallel to DF,

therefore the point F is determined.

Synthesis. Bisect the base BC in F, join DF,

from A, di-aw ^i^ parallel to DF, and join DF.

Then because DF is parallel to AF,

therefore the triangle ADF is equal to the triangle DFF; (i. 37.)

to each of these equals, add the triangle BDF,

therefore the whole triangle ABF is equal to the whole DBF,

but ABF is half of the whole triangle ABC-,

therefore DBF is also half of the triangle ABC,

PROPOSITION VII. THEOREM.

If from a point without a parallelogram lines be draioii to the extremities

of two adjacent sides, and of the diagonal ichich they include ; of the tri-

angles thus formed, that, whose base is the diagonal, is equal to the sum of

the other two.

Let ABCD be a parallelogram of which AC is one of the diagonals,

and let P be any point without it: and let AF, PC, BP, PD be

joined.

Then the triangles APD, APB are together equivalent to the tri-

angle A PC.

E

74 GEOMETRICAL EXERCISES

Draw PGE parallel to AD or BC, and me*eting ^ J? in (r, and DC

in Â£; and join DG, GC.

Then the triangles CBP, CBG are equal: (l. 37.)

and taking the common part CBH from each,

the remainders PHB, CHG are equal.

Again, the triangles DAP, DA G are equal ; (l. 37.)

also the triangles DAG, AGO are equal, being on the same base

AG, and between the same parallels AG, DC:

therefore the triangle DAP is equal to the triangle AGO:

but the triangle PHB is equal to the triangle CHG,

wherefore the triangles PUB, DAP are equal to AGC, CHG, or

A CH, add to these equals the triangle APH,

therefore the ti-iangles APH, PHB, DAP are equal to APH, ACH,

that is, the triangles APB, DAP are together equal to the triangle

PAC.

If the point P be within the parallelogram, then the difference of

the triangles APB, DAP may be proved to be equal to the triangle

PAC.

8. Describe an isosceles triangle upon a given base and having

each of the sides double of the base, without using any i^roposition of

the Elements subsequent to the first three. If the base and sides be

given, what condition must be fulfilled with regard to the magnitude

of each of the equal sides in order that an isosceles triangle may be

constructed ?

9. In the fig. Euc. I. 5. If FC and BG meet in H, then prove

that ^JJ bisects the angle BA C.

10. In the fig. Euc. i. 5. If the angle FBG be equal to the angle

ABC, and BG, CF, intersect in O; the angle BOF is equal to twice

the angle BA C.

11.' From the extremities of the base of an isosceles triangle straight

lines are di-awn perpendicular to the sides, the angles made by them

with the base are each equal to half the vertical angle.

12. A line drawn bisecting the angle contained by the two equal

sides of an isosceles triangle, bisects the third side at right angles.

13. If a straight line drawn bisecting the vertical angle of a tri-

angle also bisect the base, the triangle is isosceles.

ON BOOK 1. 75

14. Given two points one on eadi side of a given straight line ;

. find a point in the line such that the angle contained by two lines

drawn to the given points may be bisected by the given line.

15. In the fig. Euc. I. 5, let F and G be the points in the sides

AB and reproduced, and let lines FII and 6r^be drawn perpen-

dicular and equal to FC and GB respectively : also if BH, CK, or

these lines produced meet in O ; prove that BH is equal to CK, and

BO to CO.

16. From every point of a given straight line, the straight linej;

drawn to each of two given points on opposite sides of the line are

equal : prove that the line joining the given points will cut the given

line at right angles.

17. If r be the vertex of an isosceles triangle ABC, and BA be

produced so that AD is equal to BA, and DC be drawn; shew thai

B CD is a right angle.

18. The straight line EDF, drawn at right angles to J5Cthe base

of an isosceles triangle ABC, cuts the side AB in D, and CA pro-

duced in E; shew that AED is an isosceles triangle.

19. In the fig. Euc. I. 1, if AB be produced both ways to meet

the circles in D and E, and from C, CD and CE be drawn ; the figure

CDE is an isosceles triangle having each of the angles at the base,

equal to one fourth of the angle at the vertex of the triangle.

20. From a given point, draw two straight lines making equal

angles with two given straight lines intersecting one another.

21. From a given point to draw a straight line to a given straight

line, that shall be bisected by another given straight line.

22. Place a straight line of given length between two given

straight lines which meet, so that it shall be equally inclined to each

of them.

23. To determine that point in a straight line from which the

straight lines di^awn to two other given points shall be equal, pro-

vided the line joining the two given points is not perpendicular to the

given line.

24. In a given straight line to find a point equally distant from

two given straight lines. In what case is this impossible ?

25. If a line intercepted between the extremity of the base of an

isosceles triangle, and the opposite side (produced if necessary) be

equal to a side of the triangle, the angle formed by this line and the

base produced, is equal to three times either of the equal angles of the

tiiangle.

26. In the base J5C of an isosceles triangle ABC, take a point D,

and in CA take CE equal to CD, let ED produced meet reproduced

in F, then '3.AEF= 2 right angles + AFE, or = 4 right angles + AFE.

27. If from the base to the opposite sides of an isosceles triangle,

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38