Euclid.

# Euclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools online

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Font size 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

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 ÂŁ; 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,