Lawrence S. (Lawrence Sluter) Benson. # Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson online

. **(page 2 of 21)**

Online Library → Lawrence S. (Lawrence Sluter) Benson → Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson → online text (page 2 of 21)

Font size

Col. P. Thomson's Oeometry xoithout Axioim, Professors Thomson's and

Simson's editions of Euclid â€” London, Glasgow, and Belfast.

f See fifteenth and nineteenth propositions cf this book.

14 THE ELEMENTS OF [bOOK I.

POSTULATES,

1. That a straight line can be drawn from any one point to

any other point.

2. That a terminated straight line may be extended to any

length in a straight line.

3. That a circle may be described from any center, at any

distance from that center.

EXPLAIS'ATION OF SIGSTS.

In Algebra, the sign +, called Plus {more Jy), placed between

the names of two magnitudes, is used to denote that these mag-

nitudes are added together ; and the sign â€” , called Minus

{less hy)^ placed between them, to signify that the latter is taken

from the former. The sign z=, which is read equal to, signifies

that the quantities between which it stands are equal to one

another. The sign =o=, signifies that the quantities between

which it stands are equivalent to one another.

In the references, the Roman numerals denote the book, and

the others, when no word is annexed to them, indicate the

proposition ; otherwise the latter denote a definition, postulate,

or axiom, as specified. Thus, III. 16 means the sixteenth

proposition of the third book ; and I. ax. 2, the second axiom

of the first book. So also hyp. denotes hypothesis, and const,

construction.

PROPOSITIONS.

Prop. I. â€” Problem. â€” To describe an isosceles triangle on a

finite straight line given in position.

Let AX be the given straight line; it is required to describe

an isosceles triangle having its base on AX.

From a point C, without the line AX as

a center, and a radius CA (I. post. 3), de-

scribe a circle ABED, cutting the line AX

in two points A and B; draw from these

points the straight lines AE and BD (I.

post. 1) passing through the center of the

circle ; the triangle ACB is the one required.

Because C is the center of the circle ABED (I. def 16), CA

BOOK I.J EUCLID AND LEGENDRE. 15

is equal to CB, therefore the triangle ACB has two sides equal ;

hence (I. def. 13) it is isosceles, and is described on AX, which

was required to be done.

Corollary 1. But the angle EAB is subtended by the arc

EB (I. def. 19), and the angle DBA is subtended by the arc

DA ; since AE and DB pass through the center of the circle

(const,), they are both diameters of the circle (I. def 17) ; hence

the arcs DEB and ADE are each a semicii-cumference, and (I.

ax. 1) are equal ; therefore the sum of tlie arcs BE and ED

is equivalent to the sum of the arcs ED and DA ; the arc ED

is common ; hence (I. ax. 3) we have the arc EB equal to the

arc DA ; therefore the angles EAB and DBA are subtended

by equal arcs, consequently (I. def 19) the angles are equal.

Hence, in an isosceles triangle, the angles opposite the equal

sides are equal.

Cor. 2. The line AE, which forms with AB the angle EAB,

intercepts the line DB at C, which forms with AB the angle

DBA, and the line DB intercepts AE at C also. C being the

center of the circle ABED, CB, that portion of BD intercepted

hy AE, is equal to CA, that portion of AE intercepted by BD

(I. def 16) ; but CB and CA are the sides of the triangle ACB

(I. 1) ; hence, when two angles of a triangle are equal, the

opposite sides to them are also equal, and the triangle is isos-

celes (I. def 13).

Pkop. II. â€” Problem. â€” To describe an equilateral triangle

on a finite straight line given in magnitude.

Let AB be the given straight line ; it is required to describe

an equilateral triangle having AB for its base.

From A as a center, and a radius

AB (I. post. 3), describe the circle

FCD ; and from B as a center, and

a radius BA, describe the circle

HCE. The circles having equal

radii (I. ax. 1) are equal ; draw from

C through the center B, CH ; and

from C through the center A, CF (I. post 1) ; the triangle ACB

is the one required.

Because the circles have equal radii (const.), AC is equal to

16 THE ELEMENTS OF [bOOK I.

AB, and CB is equal to AB ; hence (I. ax. 1) the three sides

of the triangle ACB are equal; the triangle (I. def. 13) is equi-

lateral, and is described on AB, which was required to be done.

Corollary 1. If AB be produced both ways (I. post. 2) to D

and E, the angle DBC is subtended by the arc CD, and the

angle FCB is subtended by the arc FB; the arcs OB and BF

are together equivalent to the arcs BC and CD (I. ax. 1) ;

hence (I. 1, cor. 1) the arc BF is equal to the arc CD, therefore

(I. def 19) the angle FCB is equal to the angle DCB. Again:

the angle ACH is subtended by the arc AH, and the angle

CAE is subtended by the arc CE. But (I. ax. 1) the arcs AC

and CE are equivalent to the arcs CA and AH, and (I. ax. 3)

the arcs CE and AH are equal; therefore (I. def 19) the angles

ACH and CAE are equal, but the angle ACH is the same as

the angle FCB ; hence (I. ax. 1) the three angles of the triangle

are equal ; therefore in an equilateral triangle the angles are

equal. And in a manner similar to Cor. 2 of the first proi)osi-

tion,it can be shown, conversely^ that when a triangle has three

equal angles, the sides opposite them are also equal ; hence an

equilateral triangle is also equiangular, and, conversely^ an equi-

angular triangle is also equilateral.

Prop. IH. â€” Theorem. â€” If two triangles have txoo sides of

the one equal to two sides of the other, each to each, and have

also an angle in one equal to an angle in the other simiiiarly

situated with respect to those sides, the triangles have their

bases or remaining sides equal / their other angles equal, each

to each, viz., those to which the equal sides a.e opposite, and

the triangles are equal.

This general proposition has four cases, viz. : first, when the

equal angles are contained by the respectively equal sides;

Becond, when the equal angles are opposite to one pair of the

respectively equal sides ; third, when the equal angles are op-

posite to the other pair of the respectively equal sides ; and

fourth, the limitation that when the least sides respectively of

the triangles be equal, and the angles opposite the least sides

"be equal, the angles opposite the greater of the respectively

equal sides must be of the same kind, either both acute, or not

acute.

BOOK I.]

EUCLID AND LEGENDRE.

17

First case. Let ABC and DEF be the two triangles having

any two sides equal, each to each,

viz., AC and CB equal to EF and

DF, and the contained angles ACB

and EFD equal ; the remaining sides

AB and DE are equal, tlie angle CBA

opposite AC equal to the angle FDE

opposite FE, the angle CAB opposite

CB equal to the angle FED opposite

DF, and tlie triangles ABC and DEF

are equal.

If the triangle ABC be placed on the triangle DEF so that

the vertex of the angle ACB will fall on the vertex of the

angle DFE, the angle ACB being equal to the angle DFE

(hyp.), the side CB will fall on FD, and the side CA will fall on

FE ; CB and FD being equal (hyp.), the extremity B will fall

on the extremity D. CA and FE being equal (hyp.), the ex-

tremity A will fall on the extremity E ; and since AB is a

straight line, it will coincide with DE (I. def. V), a straight line

drawn from D to E. Therefore the triangle ABC has its three

sides coinciding with the three sides of the triangle DEF ;;

hence the angle CAB will fall on the angle FED, and be equal

to it ; the angle CBA will fall on the angle FDE, and be equal

to it ; consequently the two triangles have their three sides and.

three angles equal, each to each, and (I. ax. 8) are equal.

Second case. When the triangles ABC and DEF have the

sides CA and CB respectively equal to FE and FD, and the

angles ABC and EDF equal, respectively opposite to CA and

FE, the remaining sides are equal ; the angle CAB opposite CB

is equal to the angle FED opposite to FD, the angle ACB op-

posite to AB is equal to the angle EFD opposite to DE, and

the triangles are equal.

Let the side DE b

be put on AB so

that D will fall

on B, and the

eqnal angles ABC

and EDF will be

on different sides of AB ; join CF (I. post. 1 ). BC and BF (hyp.) are

2

18 THE ELEMENTS OF [bOOK I.

equal, the triangle CBF (I. def. 13) is isosceles, and (I. 1, cor. l)

the angle BCF eqiial to the angle BFC; and hecause CA and,

AF are equal (hyp.), the triangle CAF (I. def. 13) is isosceles,

and (1, 1, cor. 1) the angle ACF is equal to the angle AFC ;

then (I. ax. 2) the angles BCF and ACF are equal to the angles

BFC and AFC, or the angle BCA equal to the angle BFA

(I. ax. 1 and ax. 10). Hence we have in the triangles ABC

and ABF, two sides, and the contained angle in each equal,

.-Â«ach to each, therefore hj first case the triangles can be shown

.-.equal in all respects.

Third case. It can be proven in a similar manner as the second

* case. When the angles CAB and FED of the second case are

, obtuse, and the angles CBA and FDE of the third case are

. obtuse, the proofs are given by the third axiom of the first book.

Fourth case. When the triangles ABC and DEF have

their least sides in each equal â€” viz., AB to DE â€” and another

side in each, equal, the angles ACB and EFD being equal, the

angles opposite the second pair of equal sides must be both

acute or both not acute ; otherwise, two triangles can be formed

having two sides .and an augl'-^ in each equal, each to each, and

? the triangles unequal. For, in the

triano-lcs ABC and ACD, the side

AC is common, the angle BAC

equal to angle CAD, the sides BC

and CD can be equal, and the tri-

angles (I. ax. 9) unequal; hence in two triangles when the

greatest and least sides are respectively equal, and the equal

angles opposite to the least sides be given, the angles opposite

the greatest sides must both be not acute to determine the tri-

angles ; but when in two triangles the two less sides of each

are respectively equal, and the equal angles opposite tlie least

sides be given, the angles opposite the other equal sides must

both be acute, to determine the triangles.

The equality of the triangles can be proven by {ho second

and third cases, using the second axiom when the angles are

acute, and the third axiom when the angles are obtuse ; but

when the angles are right-angled, the equality of the triangles

is shown from tlie first corollary to the first proposition without

those axioms.

!BOOK I.]

EUCLID AND LEGENDKE.

19

B

Peop. IV. â€” Theok. â€” If the three, sides of one triangle be

â– equal to the three sides of another, each to each: (1) the ajigles

â– of one triayigle are equal to the angles of the other, each to each^

viz., those to which tJie equal sides are 02)posite, and (2) the

triangles are equal.

Let ABC and DEF be the two tri- c p

angles having then- three sides equal,

viz., AB to DE, CA to FE, and CB

to FD, the angles are equal, viz., ACB

to EFD, CAB to FED, and CBA

to FDE ; and the triangles are equal.

If the side DE be placed on the

side AB so that the triangles Avill

fall on different sides of AB, D will

fall on B, and E on A, because DE is equal to AB, and the

triangle DEF will

take the position

BFA, BF being

the same as DF,

and FA the same

as FE. Join CF, a

and because (hyp.) BC is equal to BF, the angles BCF and

BFC are equal (I. 1, cor. 1). It would be shown in a similar

manner that the angles FCA and CFA are equal, therefore (I.

ax. 2) the angles BCA and BFA are equal â€” that is (I. ax. 1), the

angles BCA and DFE are equal. But (hyp.) the sides CB and

FD are equal, and the sides CA and FE, and it has been shown

that the contained angles are equal, therefore (I. 3, first case)

the other angles are equal â€” that is, CAB to FED and CBA to

FDE, and the triangles are equal. Wherefore, if the three

sides, etc.

Prop. V. â€” Prob. â€” To bisect a given angle, that is, to divide

it into tico equal angles.

Let BAC be the given angle ; it is required to bisect it.

From A as a center, and AD less than AB (I. post. 3), de-

scribe the arc DE ; draw the chord DE, then upon DE, on the

side remote from A, describe an equilateral triangle (I. 2), DFE,

then join AF ; AF bisects the angle BAC.

20

THE ELEMENTS OF

[book I.

Because AD is equal to AE (L def. 16),

and AF is comraoii to the two triangles

DAF and EAF, the two sides DF and

EF are equal (I. 2) ; therefore the two tri-

angles DAF and EAF have their three

sides equal, each to each, and the triangles

are equal (I. 4) ; consequently the angle

DAF opposite DF is equal to the angle

EAF opposite EF, and the angle BAG

is bisected by the line AF; which was to be

done.

OTHERWISE,

Let BAG be the given angle ; in AB take any two points as

B and D, and cut off AG and AE respectively equal to AB and

AD, join BE and GD, and the straight line joining the inter-

section of BE and CD with the vertex A bisects BAG. The

proof is easy, and is omitted to exercise the ingenuity of the

pupU.

Prop. VI. â€” Prob. â€” To bisect a given finite straight line.

Let AB be the given line ; it is required to bisect it.

Describe (L 2) upon it an equilateral triangle ABG, and

C bisect (I. 5) the angle AGB by the straight

line GD ; AB is bisected in the point D.

Because AG is equal to GB, and CD

common to the two triangles AGD, BGD,

the two sides AG, GD are equal to BG,

B CD, each to each ; and the angle AGD is

equal (const.) to the angle BGD; therefore the base AD is

equal (L 3) to the base DB, and the line AB is bisected in the

point D ; which was to be done.

Sc/io. In practice, the construction is effected more easily by

describing arcs on both sides of AB, from A as a center, and

"with any radius greater than the half of AB ; and then, by de-

gcribing arcs intersecting them, with an equal radius, from B as

center, the line joining the two points of intersection will bi-

sect AB. The proof is easy.

Peop. VII. â€” Peob. â€” To draw a straight litie perpendieuiar

BOOK I.]

EUCLID AND LKGENDRE.

21

to a given straight line, from a giveyi point in that straight

line.

Let AB be the given straight line, and C a point given in it ;

it is required to draw a perpendicular from the point C.

From C as a center, and a radius

CE (I. post 3), describe the semicircle

EHF; then (I. def. 16) EC is equal to

CF, and on EF (I. 2) describe the ^

equilateral triangle EDF ; then a line

from C to the vez'tex D is the perpen-

dicular required.

Because EC is equal to CF (I. def.

16), ED is equal to DF (I. 2), and the angle DEC equal to

angle DFC (I. 2, cor. 1) ; hence (I. 3) the triangles ECD and

FCD are equal ; but the angle ECD is equal to the angle FCD,

and therefore (I, def 1 0) DC is perpendicular to AB from C.

Pkop. Vin. â€” Prob. â€” To draw a straight line perpendicular

to a given straight line of an unlimited length, from a given

â€¢point without it.

Let AB be the given straight line, which may be produced

any length both ways, and let C be a point without it. It is

required to draw a straight line from C

perpendicular to AB.

Take any point D upon the other side

of AB, and from the center C, at the dis-

tance CD, describe (L post. 3) the circle

ADB meeting AB in A and B ; bisect (I. 6) AB in G, and join

CG ; the straight line CG is the perpendicular required.

Join CA, CB. Then, because AG is equal to GB, and CG

common to the triangles AGC, BGC, the two sides AG, GC

are equal to the two, BG, GC, each to each ; and the base CA

is equal (I. def 1 6) to the base CB ; therefore the angle CGA

is equal (I. 4) to the angle CGB ; and they are adjacent angles ;

therefore CG is perpendicular (I. def 10) to AB. Hence, from

the given point C a perpendicular CG has been drawn to the

â€¢given line AB ; which was to be done.

Scho. This proposition and the preceding contain the only

.two distinct cases of drawing a perpendicular to a given straight

22 THE ELEMENTS OF [BOOK B-

line through a given point ; the first, when the point is in the

line ; the second, when it is zcithout it.

In practice, the construction will be made rather more simple

by describing from A and B, when found, arcs on the remote

side of AB from C, with any radius greater than the half of

AB, and joining their point of intersection with C.

Prop. IX. â€” T^eor. â€” "When one straight line meets another

straight line ayid forms two unequal angles on the same side

of that line, the two angles will be equivalent to two right

angles.

^ Let the straight line DC meet the

straight line AB and form the two un-

equal angles DCA and DCB on the same

side of AB ; the two angles will be equiva-

c lent to two right angles.

At C, where the line DC meets AB, draw a perpendicular to

AB from C (I. '7), then the angles ACE and ECB are two right

angles (I. def. 10). But (I. ax. 10) the angle ECB is equivalent

to the angles ECD and DCB both together ; likewise the angles

ACE and ECB together are equivalent to the angles ACD and

DCB both together; hence (I. ax. 1) the angles ACD and DCB

together ai-e equivalent to two right angles. Wherefore, when

one straight line meets, etc.

Cor. Hence, if the straight line DC be pi-oduced on the other

side of AB, the four angles made by DC produced and AB

are together equivalent to four right angles.

Hence, also, all the angles formed by any number of straight

lines intersecting one another in a common point are together

equivalent to four right angles.

Prop. X. â€” Theor. â€” 7)^, at a point in a straight line, tico

other straight lines on the ojyposite sides make the adjacent

angles together equivalent to two right atigles, those two straight

lines are in one and the same straight line.

Let DC be the straight line which makes, at the point C,

with AC and CB, two adjacent angles ACD and DCB togethei-

equivalent to two right angles ; AC and CB are in one and the-

same straight line.

â€” B

BOOK I.] EUCLID AND LEGENDEE. 23

From C draw a perpendicular to AC (I. ^

7), then the angle ACE is a right angle

(I. def. 10). But (hyp.) ACD and DCB

are together equivalent to two right

angles, so ACE and ECB are equivalent *

to two right angles (I. ax, 1) ; hence ACE being a right angle

(const, and I. def. 10), ECB must also be a right angle; then

EC is perpendicular to CB (I. def. 10), and the angles ACE and

ECB are equal (I. ax. 11) ; therefore (I. def 10) EC is a straight

line which makes two equal angles with AB. But AC and CB

make with EC the same equal angles ; hence AC and CB are

the same straight line with AB. And DC makes with AC and

CB (hyp.) two adjacent angles equivalent to two right angles,

but DC makes with AB (I. 9) the same angles equivalent to

two right angles ; hence AC and CB are the same straight line

with AB (I. def 7).

OTIIER^VISE,

It is proven (I. 9) that the angles ACD and DCB are to-

gether equivalent to two right angles ; then, as C is a point in

AB, AC can be one line and CB another (I. def 7, scho.) ;

hence AC and CB are in one and the same straight line, be-

cause the two unequal angles ACD and DCB are equivalent to

two right angles (I. 9). Wherefore if, at a point, etc.

Prop. XI. â€” Theoe. â€” If two straight lines cut one another^

the vertical or opposite angles are equal.

Let the two straight lines AB, CD cut one another in E ; the

angle AEC is equal to the angle DEB, and CEB to AED.

Because the straight line AE makes with CD the angles

CEA, AED, these angles are together equivalent (I. 9) to two

risrht angles. Again : because DE n

makes with AB the angles AED, DEB

these also are together equivalent to

two right angles; and CEA, AED ^

have been demonstrated to be equivalent to two right angles ;

wherefore (I. ax. 11 and 1) the angles CEA, AED are equal to

the angles AED, DEB. Take away the common angle AED,

and (I. ax. 3) the remaining angles CEA, DEB are equal

24 THE ELEMENTS OF [bOOK I.

lu the same manner it can be demonstrated that the angles

CEB, AED are equal. Therefore, if two straight lines, etc.

Cor. If at a point in a straight line two other straight lines

meet on the opposite sides of it, and make equal angles with the

parts of it on opposite sides of the point, the two straight lines

are in one and the same straight line.

Let AEB be a straight line, and let the angles AEC, BED be

equal, CE, ED are in the same straight line. For, by adding

the angle CEB to the equal angles AEC, BED, we have BED,

BEC together equal to AEC, CEB, that is (I. 9), to two right

angles ; and therefore, by this proposition, CE, ED are in the

same straight line.

Scho. In the proof here given, the common angle is AED ;

and CEB might with equal propriety be made the common

angle. In like manner, in proving the equality of CEB and

AED, either AEC or BED may be made the common angle.

It is also evident, that when AEC and BED have been proved

to be equal, the equality of AED and BEC might be inferred

from the ninth proposition, and the third axiom.

Prop. XIL â€” Pkob. â€” To describe a triangle of which the sides

shall be equal to three given straight lines; but any txoo of

these must be greater than the third.

Let A, B, C be three given straight lines, of which any

two are greater than the third ; it is required to make a tri-

angle of which the sides shall be equal to A, B, C, each to each.

Take an unlimited straight line DE, and let F be a point in

it, and make FG equal to A, FH to B, and HK to C. From

the center F, at the distance FG,

describe (I. post. 3) the circle

GLM, and from the center H, at

-E the distance HK, describe the

circle KLM. Now, because (hyp.)

g FK is greater than FG, the cir-

" ^ cumference of the circle GLM

will cut FE between F and K, and therefore the circle KLM

can not lie wholly within the circle GLM. In like manner, be- ,

cause (hyp.) GH is gi-onter than HK, the circle GLM can not

lie wholly witluu the ciicle KLM. Neither can the circles be

BOOK I.] ETTCLID AND LEGENDRE. 25

wholly without each other, since (hyp.) GF and HK are to-

o-ether jrreater than FH, The circles must therefore intersect

each other ; let them intersect in the point L, and join LF, LH ;

the triangle LFII has its sides equal respectively to the three

lines A, B, C.

Because F is the center of the circle GLM, FL is equal (I.

def. 16) to FG; but (const.) FG is equal to A; therefore (I. ax.

1) FL is equal to A, In like manner it may be shown that

HL is equal to C, and (const.) FH is equal to B ; therefore the

three straight lines LF, FH, HL are respectively equal to the

three lines A, B, C ; and therefore the triangle LFH has been

constructed, having its three sides equal to the three given lines,

A, B, C ; which was to be done.

Scho. It is evident that if MF, MH were joined, another tri-

angle would be formed, having its sides equal to A, B, C, It

is also obvious that in the practical construction of this problem,

it is only necessary to take with the compasses FH equal to B,

and then, the compasses being opened successively to the

lenorths of A and C, to describe circles or arcs from F and H as

centers, intersecting in L ; and lastly to join LF, LH.

The construction in the proposition is made somewhat differ-

ent from that given in Simson's Euclid, with a vicAv to obviate

objections arising from the application of this proposition in the

one that follows it.

Prop. XIII. â€” Prob. â€” At a given point in a given straight

line, to make a rectilineal angle equal to a given one.

Let AB be the given straight line, A the given point in it,

and C the given angle ; it is required to make an angle at A,

in the straight line AB, that shall be equal to C.

In the lines containing the angle C,

take any points D, E, and join them,

and make (L 12) the triangle AFG,

the sides of which, AF, AG, FG, shall

be equal to the three straight lines

CD, CE, DE, each to each. Then,

because FA, AG are equal to DC,

CE, each to each, and the base FG to

the base DE, the angle A is equal (I.

26

THE ELEMENTS OF

[book

4) to the angle C. Therefore, at the given jsoint A in the given

straight line AB, the angle A is made equal to the given angle

C ; which was to be done.

Scho. The construction is easy, by making the triangles isos-

celes. In doing this, arcs are described with equal radii from C

and A as centers, and their chords are madÂ« equal.

It is evident that another angle might be made at A, on the

other side of AB, equal to C.

Prop. XIV. â€” Theor. â€” If two angles of one triangle he equal

to two angles of another^ each to each, and if a side of the one

he equal to a side of the other siniiliarly situated with respect

to those angles ; (l) the remaining sides are equal, each to each ;

(2) the remaining angles are equal ; and (3) the triangles are

Simson's editions of Euclid â€” London, Glasgow, and Belfast.

f See fifteenth and nineteenth propositions cf this book.

14 THE ELEMENTS OF [bOOK I.

POSTULATES,

1. That a straight line can be drawn from any one point to

any other point.

2. That a terminated straight line may be extended to any

length in a straight line.

3. That a circle may be described from any center, at any

distance from that center.

EXPLAIS'ATION OF SIGSTS.

In Algebra, the sign +, called Plus {more Jy), placed between

the names of two magnitudes, is used to denote that these mag-

nitudes are added together ; and the sign â€” , called Minus

{less hy)^ placed between them, to signify that the latter is taken

from the former. The sign z=, which is read equal to, signifies

that the quantities between which it stands are equal to one

another. The sign =o=, signifies that the quantities between

which it stands are equivalent to one another.

In the references, the Roman numerals denote the book, and

the others, when no word is annexed to them, indicate the

proposition ; otherwise the latter denote a definition, postulate,

or axiom, as specified. Thus, III. 16 means the sixteenth

proposition of the third book ; and I. ax. 2, the second axiom

of the first book. So also hyp. denotes hypothesis, and const,

construction.

PROPOSITIONS.

Prop. I. â€” Problem. â€” To describe an isosceles triangle on a

finite straight line given in position.

Let AX be the given straight line; it is required to describe

an isosceles triangle having its base on AX.

From a point C, without the line AX as

a center, and a radius CA (I. post. 3), de-

scribe a circle ABED, cutting the line AX

in two points A and B; draw from these

points the straight lines AE and BD (I.

post. 1) passing through the center of the

circle ; the triangle ACB is the one required.

Because C is the center of the circle ABED (I. def 16), CA

BOOK I.J EUCLID AND LEGENDRE. 15

is equal to CB, therefore the triangle ACB has two sides equal ;

hence (I. def. 13) it is isosceles, and is described on AX, which

was required to be done.

Corollary 1. But the angle EAB is subtended by the arc

EB (I. def. 19), and the angle DBA is subtended by the arc

DA ; since AE and DB pass through the center of the circle

(const,), they are both diameters of the circle (I. def 17) ; hence

the arcs DEB and ADE are each a semicii-cumference, and (I.

ax. 1) are equal ; therefore the sum of tlie arcs BE and ED

is equivalent to the sum of the arcs ED and DA ; the arc ED

is common ; hence (I. ax. 3) we have the arc EB equal to the

arc DA ; therefore the angles EAB and DBA are subtended

by equal arcs, consequently (I. def 19) the angles are equal.

Hence, in an isosceles triangle, the angles opposite the equal

sides are equal.

Cor. 2. The line AE, which forms with AB the angle EAB,

intercepts the line DB at C, which forms with AB the angle

DBA, and the line DB intercepts AE at C also. C being the

center of the circle ABED, CB, that portion of BD intercepted

hy AE, is equal to CA, that portion of AE intercepted by BD

(I. def 16) ; but CB and CA are the sides of the triangle ACB

(I. 1) ; hence, when two angles of a triangle are equal, the

opposite sides to them are also equal, and the triangle is isos-

celes (I. def 13).

Pkop. II. â€” Problem. â€” To describe an equilateral triangle

on a finite straight line given in magnitude.

Let AB be the given straight line ; it is required to describe

an equilateral triangle having AB for its base.

From A as a center, and a radius

AB (I. post. 3), describe the circle

FCD ; and from B as a center, and

a radius BA, describe the circle

HCE. The circles having equal

radii (I. ax. 1) are equal ; draw from

C through the center B, CH ; and

from C through the center A, CF (I. post 1) ; the triangle ACB

is the one required.

Because the circles have equal radii (const.), AC is equal to

16 THE ELEMENTS OF [bOOK I.

AB, and CB is equal to AB ; hence (I. ax. 1) the three sides

of the triangle ACB are equal; the triangle (I. def. 13) is equi-

lateral, and is described on AB, which was required to be done.

Corollary 1. If AB be produced both ways (I. post. 2) to D

and E, the angle DBC is subtended by the arc CD, and the

angle FCB is subtended by the arc FB; the arcs OB and BF

are together equivalent to the arcs BC and CD (I. ax. 1) ;

hence (I. 1, cor. 1) the arc BF is equal to the arc CD, therefore

(I. def 19) the angle FCB is equal to the angle DCB. Again:

the angle ACH is subtended by the arc AH, and the angle

CAE is subtended by the arc CE. But (I. ax. 1) the arcs AC

and CE are equivalent to the arcs CA and AH, and (I. ax. 3)

the arcs CE and AH are equal; therefore (I. def 19) the angles

ACH and CAE are equal, but the angle ACH is the same as

the angle FCB ; hence (I. ax. 1) the three angles of the triangle

are equal ; therefore in an equilateral triangle the angles are

equal. And in a manner similar to Cor. 2 of the first proi)osi-

tion,it can be shown, conversely^ that when a triangle has three

equal angles, the sides opposite them are also equal ; hence an

equilateral triangle is also equiangular, and, conversely^ an equi-

angular triangle is also equilateral.

Prop. IH. â€” Theorem. â€” If two triangles have txoo sides of

the one equal to two sides of the other, each to each, and have

also an angle in one equal to an angle in the other simiiiarly

situated with respect to those sides, the triangles have their

bases or remaining sides equal / their other angles equal, each

to each, viz., those to which the equal sides a.e opposite, and

the triangles are equal.

This general proposition has four cases, viz. : first, when the

equal angles are contained by the respectively equal sides;

Becond, when the equal angles are opposite to one pair of the

respectively equal sides ; third, when the equal angles are op-

posite to the other pair of the respectively equal sides ; and

fourth, the limitation that when the least sides respectively of

the triangles be equal, and the angles opposite the least sides

"be equal, the angles opposite the greater of the respectively

equal sides must be of the same kind, either both acute, or not

acute.

BOOK I.]

EUCLID AND LEGENDRE.

17

First case. Let ABC and DEF be the two triangles having

any two sides equal, each to each,

viz., AC and CB equal to EF and

DF, and the contained angles ACB

and EFD equal ; the remaining sides

AB and DE are equal, tlie angle CBA

opposite AC equal to the angle FDE

opposite FE, the angle CAB opposite

CB equal to the angle FED opposite

DF, and tlie triangles ABC and DEF

are equal.

If the triangle ABC be placed on the triangle DEF so that

the vertex of the angle ACB will fall on the vertex of the

angle DFE, the angle ACB being equal to the angle DFE

(hyp.), the side CB will fall on FD, and the side CA will fall on

FE ; CB and FD being equal (hyp.), the extremity B will fall

on the extremity D. CA and FE being equal (hyp.), the ex-

tremity A will fall on the extremity E ; and since AB is a

straight line, it will coincide with DE (I. def. V), a straight line

drawn from D to E. Therefore the triangle ABC has its three

sides coinciding with the three sides of the triangle DEF ;;

hence the angle CAB will fall on the angle FED, and be equal

to it ; the angle CBA will fall on the angle FDE, and be equal

to it ; consequently the two triangles have their three sides and.

three angles equal, each to each, and (I. ax. 8) are equal.

Second case. When the triangles ABC and DEF have the

sides CA and CB respectively equal to FE and FD, and the

angles ABC and EDF equal, respectively opposite to CA and

FE, the remaining sides are equal ; the angle CAB opposite CB

is equal to the angle FED opposite to FD, the angle ACB op-

posite to AB is equal to the angle EFD opposite to DE, and

the triangles are equal.

Let the side DE b

be put on AB so

that D will fall

on B, and the

eqnal angles ABC

and EDF will be

on different sides of AB ; join CF (I. post. 1 ). BC and BF (hyp.) are

2

18 THE ELEMENTS OF [bOOK I.

equal, the triangle CBF (I. def. 13) is isosceles, and (I. 1, cor. l)

the angle BCF eqiial to the angle BFC; and hecause CA and,

AF are equal (hyp.), the triangle CAF (I. def. 13) is isosceles,

and (1, 1, cor. 1) the angle ACF is equal to the angle AFC ;

then (I. ax. 2) the angles BCF and ACF are equal to the angles

BFC and AFC, or the angle BCA equal to the angle BFA

(I. ax. 1 and ax. 10). Hence we have in the triangles ABC

and ABF, two sides, and the contained angle in each equal,

.-Â«ach to each, therefore hj first case the triangles can be shown

.-.equal in all respects.

Third case. It can be proven in a similar manner as the second

* case. When the angles CAB and FED of the second case are

, obtuse, and the angles CBA and FDE of the third case are

. obtuse, the proofs are given by the third axiom of the first book.

Fourth case. When the triangles ABC and DEF have

their least sides in each equal â€” viz., AB to DE â€” and another

side in each, equal, the angles ACB and EFD being equal, the

angles opposite the second pair of equal sides must be both

acute or both not acute ; otherwise, two triangles can be formed

having two sides .and an augl'-^ in each equal, each to each, and

? the triangles unequal. For, in the

triano-lcs ABC and ACD, the side

AC is common, the angle BAC

equal to angle CAD, the sides BC

and CD can be equal, and the tri-

angles (I. ax. 9) unequal; hence in two triangles when the

greatest and least sides are respectively equal, and the equal

angles opposite to the least sides be given, the angles opposite

the greatest sides must both be not acute to determine the tri-

angles ; but when in two triangles the two less sides of each

are respectively equal, and the equal angles opposite tlie least

sides be given, the angles opposite the other equal sides must

both be acute, to determine the triangles.

The equality of the triangles can be proven by {ho second

and third cases, using the second axiom when the angles are

acute, and the third axiom when the angles are obtuse ; but

when the angles are right-angled, the equality of the triangles

is shown from tlie first corollary to the first proposition without

those axioms.

!BOOK I.]

EUCLID AND LEGENDKE.

19

B

Peop. IV. â€” Theok. â€” If the three, sides of one triangle be

â– equal to the three sides of another, each to each: (1) the ajigles

â– of one triayigle are equal to the angles of the other, each to each^

viz., those to which tJie equal sides are 02)posite, and (2) the

triangles are equal.

Let ABC and DEF be the two tri- c p

angles having then- three sides equal,

viz., AB to DE, CA to FE, and CB

to FD, the angles are equal, viz., ACB

to EFD, CAB to FED, and CBA

to FDE ; and the triangles are equal.

If the side DE be placed on the

side AB so that the triangles Avill

fall on different sides of AB, D will

fall on B, and E on A, because DE is equal to AB, and the

triangle DEF will

take the position

BFA, BF being

the same as DF,

and FA the same

as FE. Join CF, a

and because (hyp.) BC is equal to BF, the angles BCF and

BFC are equal (I. 1, cor. 1). It would be shown in a similar

manner that the angles FCA and CFA are equal, therefore (I.

ax. 2) the angles BCA and BFA are equal â€” that is (I. ax. 1), the

angles BCA and DFE are equal. But (hyp.) the sides CB and

FD are equal, and the sides CA and FE, and it has been shown

that the contained angles are equal, therefore (I. 3, first case)

the other angles are equal â€” that is, CAB to FED and CBA to

FDE, and the triangles are equal. Wherefore, if the three

sides, etc.

Prop. V. â€” Prob. â€” To bisect a given angle, that is, to divide

it into tico equal angles.

Let BAC be the given angle ; it is required to bisect it.

From A as a center, and AD less than AB (I. post. 3), de-

scribe the arc DE ; draw the chord DE, then upon DE, on the

side remote from A, describe an equilateral triangle (I. 2), DFE,

then join AF ; AF bisects the angle BAC.

20

THE ELEMENTS OF

[book I.

Because AD is equal to AE (L def. 16),

and AF is comraoii to the two triangles

DAF and EAF, the two sides DF and

EF are equal (I. 2) ; therefore the two tri-

angles DAF and EAF have their three

sides equal, each to each, and the triangles

are equal (I. 4) ; consequently the angle

DAF opposite DF is equal to the angle

EAF opposite EF, and the angle BAG

is bisected by the line AF; which was to be

done.

OTHERWISE,

Let BAG be the given angle ; in AB take any two points as

B and D, and cut off AG and AE respectively equal to AB and

AD, join BE and GD, and the straight line joining the inter-

section of BE and CD with the vertex A bisects BAG. The

proof is easy, and is omitted to exercise the ingenuity of the

pupU.

Prop. VI. â€” Prob. â€” To bisect a given finite straight line.

Let AB be the given line ; it is required to bisect it.

Describe (L 2) upon it an equilateral triangle ABG, and

C bisect (I. 5) the angle AGB by the straight

line GD ; AB is bisected in the point D.

Because AG is equal to GB, and CD

common to the two triangles AGD, BGD,

the two sides AG, GD are equal to BG,

B CD, each to each ; and the angle AGD is

equal (const.) to the angle BGD; therefore the base AD is

equal (L 3) to the base DB, and the line AB is bisected in the

point D ; which was to be done.

Sc/io. In practice, the construction is effected more easily by

describing arcs on both sides of AB, from A as a center, and

"with any radius greater than the half of AB ; and then, by de-

gcribing arcs intersecting them, with an equal radius, from B as

center, the line joining the two points of intersection will bi-

sect AB. The proof is easy.

Peop. VII. â€” Peob. â€” To draw a straight litie perpendieuiar

BOOK I.]

EUCLID AND LKGENDRE.

21

to a given straight line, from a giveyi point in that straight

line.

Let AB be the given straight line, and C a point given in it ;

it is required to draw a perpendicular from the point C.

From C as a center, and a radius

CE (I. post 3), describe the semicircle

EHF; then (I. def. 16) EC is equal to

CF, and on EF (I. 2) describe the ^

equilateral triangle EDF ; then a line

from C to the vez'tex D is the perpen-

dicular required.

Because EC is equal to CF (I. def.

16), ED is equal to DF (I. 2), and the angle DEC equal to

angle DFC (I. 2, cor. 1) ; hence (I. 3) the triangles ECD and

FCD are equal ; but the angle ECD is equal to the angle FCD,

and therefore (I, def 1 0) DC is perpendicular to AB from C.

Pkop. Vin. â€” Prob. â€” To draw a straight line perpendicular

to a given straight line of an unlimited length, from a given

â€¢point without it.

Let AB be the given straight line, which may be produced

any length both ways, and let C be a point without it. It is

required to draw a straight line from C

perpendicular to AB.

Take any point D upon the other side

of AB, and from the center C, at the dis-

tance CD, describe (L post. 3) the circle

ADB meeting AB in A and B ; bisect (I. 6) AB in G, and join

CG ; the straight line CG is the perpendicular required.

Join CA, CB. Then, because AG is equal to GB, and CG

common to the triangles AGC, BGC, the two sides AG, GC

are equal to the two, BG, GC, each to each ; and the base CA

is equal (I. def 1 6) to the base CB ; therefore the angle CGA

is equal (I. 4) to the angle CGB ; and they are adjacent angles ;

therefore CG is perpendicular (I. def 10) to AB. Hence, from

the given point C a perpendicular CG has been drawn to the

â€¢given line AB ; which was to be done.

Scho. This proposition and the preceding contain the only

.two distinct cases of drawing a perpendicular to a given straight

22 THE ELEMENTS OF [BOOK B-

line through a given point ; the first, when the point is in the

line ; the second, when it is zcithout it.

In practice, the construction will be made rather more simple

by describing from A and B, when found, arcs on the remote

side of AB from C, with any radius greater than the half of

AB, and joining their point of intersection with C.

Prop. IX. â€” T^eor. â€” "When one straight line meets another

straight line ayid forms two unequal angles on the same side

of that line, the two angles will be equivalent to two right

angles.

^ Let the straight line DC meet the

straight line AB and form the two un-

equal angles DCA and DCB on the same

side of AB ; the two angles will be equiva-

c lent to two right angles.

At C, where the line DC meets AB, draw a perpendicular to

AB from C (I. '7), then the angles ACE and ECB are two right

angles (I. def. 10). But (I. ax. 10) the angle ECB is equivalent

to the angles ECD and DCB both together ; likewise the angles

ACE and ECB together are equivalent to the angles ACD and

DCB both together; hence (I. ax. 1) the angles ACD and DCB

together ai-e equivalent to two right angles. Wherefore, when

one straight line meets, etc.

Cor. Hence, if the straight line DC be pi-oduced on the other

side of AB, the four angles made by DC produced and AB

are together equivalent to four right angles.

Hence, also, all the angles formed by any number of straight

lines intersecting one another in a common point are together

equivalent to four right angles.

Prop. X. â€” Theor. â€” 7)^, at a point in a straight line, tico

other straight lines on the ojyposite sides make the adjacent

angles together equivalent to two right atigles, those two straight

lines are in one and the same straight line.

Let DC be the straight line which makes, at the point C,

with AC and CB, two adjacent angles ACD and DCB togethei-

equivalent to two right angles ; AC and CB are in one and the-

same straight line.

â€” B

BOOK I.] EUCLID AND LEGENDEE. 23

From C draw a perpendicular to AC (I. ^

7), then the angle ACE is a right angle

(I. def. 10). But (hyp.) ACD and DCB

are together equivalent to two right

angles, so ACE and ECB are equivalent *

to two right angles (I. ax, 1) ; hence ACE being a right angle

(const, and I. def. 10), ECB must also be a right angle; then

EC is perpendicular to CB (I. def. 10), and the angles ACE and

ECB are equal (I. ax. 11) ; therefore (I. def 10) EC is a straight

line which makes two equal angles with AB. But AC and CB

make with EC the same equal angles ; hence AC and CB are

the same straight line with AB. And DC makes with AC and

CB (hyp.) two adjacent angles equivalent to two right angles,

but DC makes with AB (I. 9) the same angles equivalent to

two right angles ; hence AC and CB are the same straight line

with AB (I. def 7).

OTIIER^VISE,

It is proven (I. 9) that the angles ACD and DCB are to-

gether equivalent to two right angles ; then, as C is a point in

AB, AC can be one line and CB another (I. def 7, scho.) ;

hence AC and CB are in one and the same straight line, be-

cause the two unequal angles ACD and DCB are equivalent to

two right angles (I. 9). Wherefore if, at a point, etc.

Prop. XI. â€” Theoe. â€” If two straight lines cut one another^

the vertical or opposite angles are equal.

Let the two straight lines AB, CD cut one another in E ; the

angle AEC is equal to the angle DEB, and CEB to AED.

Because the straight line AE makes with CD the angles

CEA, AED, these angles are together equivalent (I. 9) to two

risrht angles. Again : because DE n

makes with AB the angles AED, DEB

these also are together equivalent to

two right angles; and CEA, AED ^

have been demonstrated to be equivalent to two right angles ;

wherefore (I. ax. 11 and 1) the angles CEA, AED are equal to

the angles AED, DEB. Take away the common angle AED,

and (I. ax. 3) the remaining angles CEA, DEB are equal

24 THE ELEMENTS OF [bOOK I.

lu the same manner it can be demonstrated that the angles

CEB, AED are equal. Therefore, if two straight lines, etc.

Cor. If at a point in a straight line two other straight lines

meet on the opposite sides of it, and make equal angles with the

parts of it on opposite sides of the point, the two straight lines

are in one and the same straight line.

Let AEB be a straight line, and let the angles AEC, BED be

equal, CE, ED are in the same straight line. For, by adding

the angle CEB to the equal angles AEC, BED, we have BED,

BEC together equal to AEC, CEB, that is (I. 9), to two right

angles ; and therefore, by this proposition, CE, ED are in the

same straight line.

Scho. In the proof here given, the common angle is AED ;

and CEB might with equal propriety be made the common

angle. In like manner, in proving the equality of CEB and

AED, either AEC or BED may be made the common angle.

It is also evident, that when AEC and BED have been proved

to be equal, the equality of AED and BEC might be inferred

from the ninth proposition, and the third axiom.

Prop. XIL â€” Pkob. â€” To describe a triangle of which the sides

shall be equal to three given straight lines; but any txoo of

these must be greater than the third.

Let A, B, C be three given straight lines, of which any

two are greater than the third ; it is required to make a tri-

angle of which the sides shall be equal to A, B, C, each to each.

Take an unlimited straight line DE, and let F be a point in

it, and make FG equal to A, FH to B, and HK to C. From

the center F, at the distance FG,

describe (I. post. 3) the circle

GLM, and from the center H, at

-E the distance HK, describe the

circle KLM. Now, because (hyp.)

g FK is greater than FG, the cir-

" ^ cumference of the circle GLM

will cut FE between F and K, and therefore the circle KLM

can not lie wholly within the circle GLM. In like manner, be- ,

cause (hyp.) GH is gi-onter than HK, the circle GLM can not

lie wholly witluu the ciicle KLM. Neither can the circles be

BOOK I.] ETTCLID AND LEGENDRE. 25

wholly without each other, since (hyp.) GF and HK are to-

o-ether jrreater than FH, The circles must therefore intersect

each other ; let them intersect in the point L, and join LF, LH ;

the triangle LFII has its sides equal respectively to the three

lines A, B, C.

Because F is the center of the circle GLM, FL is equal (I.

def. 16) to FG; but (const.) FG is equal to A; therefore (I. ax.

1) FL is equal to A, In like manner it may be shown that

HL is equal to C, and (const.) FH is equal to B ; therefore the

three straight lines LF, FH, HL are respectively equal to the

three lines A, B, C ; and therefore the triangle LFH has been

constructed, having its three sides equal to the three given lines,

A, B, C ; which was to be done.

Scho. It is evident that if MF, MH were joined, another tri-

angle would be formed, having its sides equal to A, B, C, It

is also obvious that in the practical construction of this problem,

it is only necessary to take with the compasses FH equal to B,

and then, the compasses being opened successively to the

lenorths of A and C, to describe circles or arcs from F and H as

centers, intersecting in L ; and lastly to join LF, LH.

The construction in the proposition is made somewhat differ-

ent from that given in Simson's Euclid, with a vicAv to obviate

objections arising from the application of this proposition in the

one that follows it.

Prop. XIII. â€” Prob. â€” At a given point in a given straight

line, to make a rectilineal angle equal to a given one.

Let AB be the given straight line, A the given point in it,

and C the given angle ; it is required to make an angle at A,

in the straight line AB, that shall be equal to C.

In the lines containing the angle C,

take any points D, E, and join them,

and make (L 12) the triangle AFG,

the sides of which, AF, AG, FG, shall

be equal to the three straight lines

CD, CE, DE, each to each. Then,

because FA, AG are equal to DC,

CE, each to each, and the base FG to

the base DE, the angle A is equal (I.

26

THE ELEMENTS OF

[book

4) to the angle C. Therefore, at the given jsoint A in the given

straight line AB, the angle A is made equal to the given angle

C ; which was to be done.

Scho. The construction is easy, by making the triangles isos-

celes. In doing this, arcs are described with equal radii from C

and A as centers, and their chords are madÂ« equal.

It is evident that another angle might be made at A, on the

other side of AB, equal to C.

Prop. XIV. â€” Theor. â€” If two angles of one triangle he equal

to two angles of another^ each to each, and if a side of the one

he equal to a side of the other siniiliarly situated with respect

to those angles ; (l) the remaining sides are equal, each to each ;

(2) the remaining angles are equal ; and (3) the triangles are

Online Library → Lawrence S. (Lawrence Sluter) Benson → Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson → online text (page 2 of 21)