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

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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 11 of 21)

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duplicate ratio of AB to AE ; but ABD

+ ADC=ABDC, and AEF+AFG =

AEFG ; wherefore the squares are to

one another in the duplicate ratio of

their sides.

For like reason, the, triangles AIID,

AFC are to one another in the duplicate

ratio that PH is to DF, or that AD is to AC ; hence, similar

triangles are one to another in the duplicate ratio of their alti-

tudes or bases ; since the segments AHD, AFC have the same

bases and altitudes as the triangles AHD and AFC, and being

segments of quadrants, are similar ;

hence, they are to one another in the du-

plicate ratio of their bases and altitudes ;

therefore, by composition, AED + seg.

AHD : ABC + seg. AFC in the duplicate

ratio of their homologous sides, or simi-

lar polygons are to one another in the

duplicate ratio of their homologous sides,

and quadrants of circles are one to an-

other in the duplicate ratio of their radii ; hence, semicircles

are one to another in the duplicate ratio of their diameters, and.

circles are one to another in the duplicate ratio of their diame-

ters. Wherefore, similar surfaces are, etc.

Cor. 1. Therefore, universally, if three lines be proportionals,

the first is to the third as any plane figure upon the first to a

similar and similarly described figure upon the second.

Cor. 2. Because all squares are similar figures, and all circles

are similar figures, the ratio of any two squares to one another

is the same as the duplicate ratio of their sides ; and the ratio

of any two circles to one another is the same as the duplicate

ratio of their diameters ; hence, any two similar plane figures

are to one another as the squares or circles (IV. 7) described on

their homolocrous sides.

Cor. 3. Because the sides of similar plane figures are propor-

tional, therefore (IV. 8) their perimeters or peripheries are

propoi'tional to the homologous sides ; hence, the perimeters of

128 THE ELKMENTS OF [bOOK V.

similar polyo^ons are proportional to their apotliems ; the cir-

cumfei-enees of circles are proportional to their diameters.

Cor. 4. Hence plane figures which are similar, to the same

figure are similar to one another (I. ax. l).

Cor. 5. If four straight lines be proportionals, the similar

plane figures, similarly desci-ibed upon them, are also propor-

tionals; and, conversely., if the similar plane figures, similarly-

described upon four straight lines, be proportionals, those lines

are proportionals.

Cor. 6. Similar polygons inscribed in circles are to one an-

other as the squares of the diameters.

Cor. 1. When there are three parallelograms, AC, CH, CF,

the first, AC (IV. def 10), has to the third, CF, the ratio which

is compounded of the ratio of the first, AC, to the second, CIT,

and of the ratio of CH to the third, CF; but AC is to CH as

their bases ; and CH is to CF as their bases ; therefore AC has

to CF the ratio which is compounded of ratios that are the

same with the ratios of the sides.

Scho. 3. Dr. Simson remarks in his Note on this corollaiy,

that "nothing is usually reckoned more difficult in the elements

of geometry by learners, than the doctrine of compound ratio."

This distinguished geometer, however, has both freed the text

of Euclid from the errors introduced by Theon or others, and

has explained the subject in such a manner as to remove the

difficulties that were formeily felt. According to liim, " every

proposition in which compound ratio is made use of, may with-

out it be both enunciated and demonstrated ;" and " the use of

compound ratio consists wholly in this, that by means of it, cir-

cumlocutions may be avoided, and thereby propositions may-

be more briefly either enunciated or demonstrated, or both

may be done. For instance, if this corollary were to be enun-

ciated, without mentioning compound ratio, it might be done

as follows: If two pai-allelograms be equiangular, and if as a

side of the first to a side of the second, so any assumed straight

line be made to a second straight line ; and as the other side

of the first to the other side of the second, so the second

straight line be made to a third ; the first parallelogram is to

the second as the first straight line to the third ; and the

demonstration would be exactly the same as we now have it.

BOOK v.] EUCLTD AND LEGENDKE. 129

But the ancient cceometers, when they observed this ennncia-

ti(ni could l)e made nhortev, l)y giving a name to the ratio

which the first straiglit line lias to the last, by which name the

intermediate ratios miglA likewise be signified, of the first to

the second, and of the second to the third, and so on, if there

â– were more of them, they called this ratio of the fiist to the last,

the ratio compounded of the ratios of the first to the second,

and of the second to the third straight line; that is, in the

present example, of the ratios which are the same with the

ratios of the sides."

Scho. 4. The seventh corollary will be illustrated by the fol-

lowing proposition, which exhibits the subject in a different,

and, in some respects, a preferable light :

Triangles which have an angle of the one equal to an angle

of the othei\ are proportional to the rectangles covitcined by

the sides about those angles ; and (2) equiangular parallelO'

grams are proportional to the rectangles co7itained by their ad-

jacent sides.

1. Let ABC, DBE be two triangles, having the angles ABC,

DBE equal; the first triangle is to the second as AB.BC is to

DB.BE.

Let the triangles be placed with their equal angles coinciding,

and join CD. Then (V. 1) AB is to DB as the triangle ABC

to DBC. But (V. 1, cor. 2) AB : DB :: AB.BC :DB.BC;

theiefore (IV. 7) AB.BC is to DB.BC as the triangle ABC to

DBC. Li the same maimer it would

be shown that DB.BC is to DB.BE as

the triangle DBC to DBE ; and, there-

fore, ex ceq^io^ AB.BC is to DB.BE as

the triangle ABC to DBE.

2. If parallels to BC through A and

D, and to AB through C and E were

drawn, parallelograms would be formed which would be re-

spectively double of the triangles ABC and DBE, and which

(IV. 9) would have the same ratio as the triangles; that is, the

ratio of AB.BC to DB.BE; and this proves the second part of

the proposition.

Comparing this proposition and the corollary, we see that the

ratio which is compounded of ihe ratio of the sides, is the same

9

130

THE ELEMENTS OF

[book V.

as the ratio of their rectangles, or the same (T. 23, cor. 5) as the

ratio of their products, if they he e:q)ressed in numbers. This

conclusion might also be derived from the proof given in the

text. For (const.) DC : CE : : CG : K ; whence (V. 10, cor.)

K.DC = CE.CG. But it was proved that BC : K : : AC : CF;

or (V. 1) BC.DC : K.DC : : AC : CF; or BC.DC : CE.CG ::

AC : CF; because K.DC=CE.CG.

The twelfth proposition of this book is evidently a case of

this proposition; and the fourteenth is also easily derived

from it.

Pijop. XV. â€” TriEOPw â€” The parallelograms ahout the diago-

nal of any parallelogram are similar to the whole, and to one

another.

Let ABCD be a parallelogram, and EG, HK the parallelo-

grams about the diagonal AC ; the parallelograms EG, HK are

similar to the whole parallelogram, and to one another.

Because DC, GF are parallels, the angles ADC, AGF are

(I. 16) equal. For the same reason, be-

cause BC, EF are parallels, the angles

ABC, AEF are equal ; and (T. 15, cor. 1)

each of the angles BCD, EFG is equal

to the opposite angle DAB, and there-

fore they are equal to one another ;

Avherefore, in the parallelograms, the

angle ABC is equal to AEF, and BAC common to the two tri-

angles BAC, EAF ; therefore (V. 3, cor.) as AB ; BC : : AE :

EF. And (IV. 5), because the opposite sides of parallelograms

are equal to one another, AB : AD :: AE : AG; and DC :

BC : : GF : EF; and also CD : DA : : FG : GA. Therefore

the sides of the parallelograms BD, EG about the equal angles

are proportionals; the parallelograms are, therefore (V. def l),

similar to one another. In the same manner it would be shown

that the parallelogram BD is similar to HK. Therefore each

of the parallelograms EG, HK is similar to BD. But (V. 14,

cor.) rectilineal figures which are similar to the same figure, are

similar to one another; therefore the parallelogram EG is simi-

lar to HK. Wherefore, etc.

Scho. Hence, GF : FE : : FH : FK. Therefore the sides of

BOOK v.]

EUCLID AND LEGENDRE.

131

the paralleloojrains GK and EH, about the equal angles at F,

are reciprocally proportional ; and (V. 12) these parallelograms

are equivalent ; a conclusion which agrees with the eighth

corollary to the fifteenth proposition of the first book.

Prop. XVI. â€” Prob. â€” To describe a rectllmeal figure which

shall be similar to one given rectilineal figure., and equivalent

to one another.

Let ABC and D be given rectilineal figures. It is required

to describe a figure similar to ABC, and equal to D.

Upon the straight line BC describe (II. 5, scho.) the parallelo-

gram BE equivalent to ABC ; also upon CE describe the parallel-

ogram CM equivalent to D, having the angle FCE equal to

CBL. Therefore (I. 16 and 10) BC and CF are in a straight

line, as also LE and EM. Between BC and CF find (V. 11) a

mean proportional GH, and ^

on it describe (V. 13) the

figure GHK similar, and

similarly situated, to ABC;

GHK is the figure required.

Because BC : GH : : GH :

CF, and if three straight

lines be proportionals, as the

first is to the third, so is (V.

14, cor. 2) the figure upon the first to the similar and similarly

described figure upon the second ; therefore,

as BC to CF, so is ABC to KGH ; but (V. 1)

as BC to CF, so is BE to EF ;

therefore (IV. 7) as ABC is to KGH, so is BE to EF. But

(const.) ABC is equivalent to BE ; therefore KGH is equivalent

(IV. ax. 4) to EF; and (const.) EF is equivalent to D; where-

fore, also, KGH is equivalent to D; and it is similar to ABC.

Therefore the rectilineal figure KGH has been described, simi-

lar to ABC and equivalent to D ; which was to be done.

Prop. XVII. â€” Theor. â€” If two similar parallelograms have

a comm,on angle, and be similarly situated, they are about the

same diagonal.

Let ABCD, AGEF be two similar parallelograms having a

132

THE ELEMENTS OF

[book V.

C

E

F

B

common angle CAB, tlicy will be about the same diagonal

AD. Similar parallelograms have

their sides about equal angles pro-

portional (V. def. l). Draw the

diagonals EG and CB ; hence, AB :

AG : : AC : AE ; therefore EG ia

parallel to CB (V. 3), and the angles

AEG, ACB are equal (I. 16); like-

wise the angles EGA, CBA. The

triangles E AG, FG A are equal (I. 15, cor. 5); likewise the tri-

angles CAB, DBA ; therefore AF is equal to EG, and AD is

equal to CB ; but AF is the diagonal also of AGEF, and is in

the same straight line with AD, the diagonal of ABCD.

Wherefore, if two similar parallelograms, etc.

Cor. Hence, equiangular parallelograms have to one another

the ratio which is compounded of the ratio of their sides;

hence, triangles which have one angle of the one equal, or sup-

plemental, to one angle of the other, have to one another the

ratio which is compounded of the ratio of the sides containing

those angles.

Pkop. XVIII. â€” ^Theok. â€” Of all the parallelof/rmns that can

he inscribed in any triangle^ that which is described on the half

of one of the sides as base is the greatest.

Let ABC be a triangle, having BC, one of its sides, bisected

in D; draw (I. 18) DE parallel to BA, and EF to BC ; let also

G be any other point in BC, and

describe the parallelogram GK;

FD is greater than KG.

If G be in DC, through C draw

CL parallel to BA, and produce

FE, KH, Gil as in the figure.

Then (I. 15, cor. 8) the comj>le-

ments LTI and IID are equivalent ; and since the bases CD,

DB are equal, the parallelograms ND, DK (I. 15, cor. 5) are

equivalent. To LH add ND, and to HD add DK ; then (I.

ax. 2) the gnomort MND is equivalent to the parallelogram

KG. But (L ax. 9) DL is greater than MXD ; and therefore

BOOK V.J EUCLID AND LFGENDEE. 133

FD, which (T. 15, cor. 5) is equal to DL, is greater than KG,

which is equivalent to the gnomon JNIND. -

If G were in BD, since BD is equal to DC, AE is equal (V.

2) to EC, and AF to FB; and by drawing through A a jiaral-

lel to BC, meeting DE produced, it would he proved in the

same manner that FD is greater than the inscribed ])aralk'lo-

gram applied to BG. Therefore, of all the parallelogi-ams, etc.

Cor. Since (V. lo) all parallelograms having one angle coin-

ciding with BCL, and their diagonals with CA, arc similar, it

follows from this proposition that if, on the segments of a given

straight line, BC, two parallelograms of the same altitude be

described, one of them, DL, similar to a given parallelogram,

the other, DF, will be the greatest possible when the segments

of the line are equal.

Scho. The parallelogram FD exceeds KG by the parallelo-

gram OM similar to DL or DF, and described on OH, which

is equal to DG, the difference of the bases BD and BG. Hence

we can describe parallelograms equivalent and similar to given

rectilineal figures.

The enunciation of this proposition here given is much more

simple and intelligible than that of Euclid, and the proof is

considerably shortened, Euclid's enunciation, as given by Dr.

Simson, is as follows: "Of all pai'allelograms applied to the

same straight line, and deficient by parallelograms, similar and

similarly situated to that which is described upon the half of

the line ; that which is applied to the half, and is similar to its

defect, is the greatest." It may be remaiked, that this piopo-

sition, in its simplest case, is the same as the second corollary

to the fifth proposition of the second book.

Prop. XIX. â€” ffHEOR. â€” In equal circles^ or in the same cir-

cle^ angles, whether at the centers or circumferences^ have the

same ratio as the arcs on which they stand have to one another^

so also have the sectors.

Let, in the equal circles ALB, DNE, the angles LGK, KGC,

CGB, DHN, NHM, and IMITF be at the centers, and the angles

BAG and EDF be at the circumferences, then will those angles

have to each other the same ratio as the arcs KL, KC, CB, DN,

NM, MF, and FE have to one another.

134

THE ELEMENTS OF

[book V.

Since (T. def. 19) all angles at the center of a circle are

measured by the arcs intercepted by the sides of the angles,

and all ancjles at the circum-

ference are subtended by the

arcs intercepted by the sides

of the angles, and (III. 16)

equal angles will have equal

arcs whether they be at the

center or the circumference;

hence, the same ratio which

the arcs have to each other, will the angles also have to one

another â€” that is, when the arcs be greater, the angles will be

greater; less, less ; and equal, equal.

And since the sectors are contained (III. def V) by the sides

of the anirles and the arcs, the same ratio between the sectors

will evidently exist as there is between the arcs. Wherefore,

in equal circles, etc.

Cor. Hence, conversely^ arcs of the same or equal circles will

have the same ratio as the angles or sectors which they measure

or subtend â€” when greater, greater ; less, less ; or equal, equal.

Pkop. XX. â€” ^Peob. â€” Tlie area of a regular inscribed poly '

gon^ and that of a regular circumscribed one of the same nvirv-

her of sides being given y to find the areas of the regidar in-

scribed and circumscribed polygons having double the number

of sides.

].et A be the center of the circle, BC a side of the inscribed

polygon, and DE parallel to BC, a side of the circumscribed

one. Draw the perpendicular AFG, and the tangents BH,

CK, and join BG; then BG will be a side

of the inscribed polygon of double the

number of sides; and (111.^6, cor. l) IIK is

a side of the similar cii-cumscribed one.

/ \\ // \ Now, as a like construction would be

/ x/^ \ Tinule at each of the remaining angles of

MAN ^^Ijp iwlygon, it will be sufficient to con-

sider the i)art here represented, as the tri-

angles connected with it are evidently to each other as the poly-

gons of which they are parts. For the sake of brevity, then.

n

G K

BOOK v.] EUCLID AND LEGENDKS. 135

let P denote the polygon whose side is BC, and P' that whose

side is DE; and, in like manner, let Q and Q' represent those

whose sides are BG and HK ; P and P/ therefore, are given ;

Q and Q' required.

Now (V. ]) the triangles ABF, ABG are proportional to

their bases AF", AG, as they are also to the polygons P, Q ;

therefore AF : AG : : P : Q. The triangles ABG, ADG are

likewise as their bases AB, AD, or (V. 3) as AF, AG; and

they are also as the polygons Q and P' ; therefore AF : AG : :

Q : P' ; wherefore (IV. 7)" P : Q : : Q : P' ; so that Q is a mean

proportional between the given polygons P, P'; and, re))re-

senting them by numbers, we have Q-=PP'', so that the area

of Q will be comjyuted by -tnaltlphjing P by P', and extracting

the square root of the product.

Again : because AH bisects the angle GAD, and because

the triangles AHD, AHG are as their bases, Me have Gil : HD

: : AG : AD, or AF : AB : : AHG : AHD. But we have

already seen that AF : AG :: P : Q; and therefore (IV. 7)

AHG : AHD : : P : Q. Hence (IV. 11) AUG : ADG : : P :

P-|-Q; whence, by doubling the antece-

dents, 2 AHG : ADG : : 2P : P+Q. But

D IT P K" "R

it is evident, that whatever part the tri-

angle ADG is of P^, the same part of the

polyg )n Q' is the triangle AHK, which is

double of the trianojle AHG, Hence the

last analogy becomes Q' : P^ : : 2P : P+Q.

Now (IV. 2, cor. 1) the product of the ex-

tremes is equal to the product of the means; and therefore Q'

will be computed by dividing twice the product o/'P and V by

P+Q; and the mode of finding Q has been pointed out

already.

Prop. XXI. â€” ^Theor. â€” Of regular polygons which have

eqiial perimeters., that which has the greater mimber of sides is

the greater.

Let AB be half the sides of the polygon which has the less

number of sides, and BC a perpendicular to it, which will evi-

dently pass through the center of its inscribed or circumscribed

circle; let C be that center, and join AC. Then, ACB will be

136

THE ELEMENTS OF

[book V.

the angle at tlie center subtended by the half side AB. ]\I;ike

BCD equal to the angle subtended at the center of the other

polygon by half its side, and from C as center, with CD as ra-

dius, describe an arc cutting AC in E, and CB i)roduced in F.

Then, it is plain, that the angle ACB i'* to four right angles as

AB to the common peiimeter; and four right angles are to

DCB, as the common pei-imeter to the half of

a side of the other polygon, which, for brevi-

ty, call S; then, ex mquo^ the angle ACB is

to DC]} as AB to S. But (V. 19) the angle

ACB is to DCB as the sector ECF to the

sector DCF; and consequently (IV. 1) the

sector ECF is to DCF as AB to S, and, by

division, the sector ECD is to DCF as AB â€”

S to S. Now the triangle ACD is greater

than the sector CED, and DCB is less than DCF. But (V. 1)

these triangles are as their bases AD, DB ; therefore AD has to

DB a greater i-atio than AB â€” S to S. Hence AB, the sum of

the first and second, has to DB, the second, a greater ratio than

AB, the sum of the third and fourth, has to S, the fourth ; and

therefore (IV. 2, cor. 5) 8 is greater than DB. Let then BG

be equal to S, and draw GH parallel to DC, meeting FC pro-

duced in H. Then, since the angles GHB, DCB are equal, BH

is the perpendicular drawn from the center of the polygon hav-

ing the greater number of sides to one of the sides ; and since

this is greater than BC, the like perpendicular in the other

polygon, while the perimeters are equal, it Ibllows that the area

of that which has the gi-eater number of sides is greater than

that of the other.

Prop. XXIT. â€” Theor. â€” If the diameter of a circle 1 e divided

into any two parts, AB, BC, and if semicircles, ADBjBECy be

described on opposite sides of these, the circle is divided by

tJieir arcs into ttoo figures, GDE, FED, the boundary of each <f

tchich is equal to the circumference q/'FG ; and which are such

that AC : BC : : FG : FED, and AC : AB :: Â¥G : GDE.

For (V. 14, cor. 3) the circumferences of circles, and conse-

quently the halves of their circumferences, are to one another as

their diameters ; thereiore AB is to AC as the arc ADB to

BOOK v.] EUCLID AND LEGENDRE. 137

AFC, and BC is to AC as the arc BEC to AFC. Hence (IV.

1) AC is to AC as the compound arc

ADEC to AFC; therefore ADEC is p

equal to half the circumference ;and the

entire boundaries of the fiirures GDE,

FED are each equal to the circumfer-

ence of FG.

Again (V. 14, cor, 2): circles, and

consequently semicircles, are to one an-

other as the squares of their diameters ;

therefore AC' is to AB^ as the semicircle

AFC to the semicircle ADB. Hence, since (H. 4) AC' = ABÂ»

-f2AB.BC + BC^ we find by conversion that AC is to 2AB.

BC + BC% as the semicircle AFC to the remaining space

BDAFC; whence, by inversion, 2AB.BC + BC' is to AC' as

BDAFC to the semicircle AFC. But BC' is to AC= as the

Bemicircle BEC to the semicircle AFC; and therefore (IV. 1)

2AB.BC + 2BC' is to AC as the compound figure FDE to the

semicircle AFC. But (II. 3) 2AB.BC+2BC = 2AC.BC, and

(V. 1) 2AC.BC : AC^ : : BC : ^AC. Hence the preceding

analogy becomes BC to ^AC, as FED to the semicircle, or by

doubling the consequents, and by inversion, AC to BC, as FG

to FED ; and it would be proved, in the same manner, that

AC : AB : : FG : GDE.

Cor. Hence we can solve the curious problem, in which it is

required to divide a circle into any proposed number of parts,

equal in area and boundary; as it is only necessary to divide

the diameter into the proposed number of equal parts, and to

desci'ibe semicircles on opposite sides. Then, whatever part

AB is of AC, the same part is AEG of the circle. Their bound-

aries are also equal, the boundary of each being equal to the

circumference of the circle.

Scho. Another solution would be obtained, if the circumfer-

ence were divided into the proposed number of equal parts, and

radii drawn to the points of division. This division, hoM'ever,

can be eftected only in some particular cases by means of ele-

mentary geometry.

Pkop. XXIII. â€” ProbÂ» â€” To divide a given circle ABC into

138

THE ELEMENTS OF

[book V.

any proposed number of equal parts by means of concentric

circles.

Divide the radius AD into the proposed number of equal

parts, suppose three, in the points E, F, and through these

points draw perpendiculars to AD, meeting a semicircle de-

scribed on it as diameter in G, H;

from D as center, at the distances DG,

DH, describe the circles GL, HK ; their

circumferences divide the circle into

equal parts.

Join All, DH. Then (V. 17, cor. 2)

AD, DH, DF being continual propor-

tionals, AD is to DF as a square de-

scribed on AD is to one described on

DH. But (V. 14, cor. 2) circles are proportional to the squares

of their diameters, and consequently to the squares of their

radii. Hence (IV. 7) AD is to FD as the circle ABC to the

circle HK ; and therefore, since FD is a third of AD, HK is a

third of ABC. It would be proved in a similar manner that

AD is to ED as ABC to GL. But ED is two thirds of AD,

and therefore GL is two thirds of ABC ; wherefore the space

between the circumferences of GL, HK is one third of ABC,

as is also the remaining space between the circumferences of

ABC and GL.

Cor. Hence it is plain that the area of any annulus^ or nng,

between the circumferences of two concentric circles, such as

that between the circumferences of ABC and GL is to the cir-

cle ABC as the difference of the squares of the radii DM, DL

to the square of DM ; or (II. 5, cor. 1, and HI. 20) as the rect-

angle AL.LM, or the square of the perpendicular LB to the

square of DM; and it therefore follows (V. 14) that the ring is

equivalent to a circle described with a radius equal to LB.

Prop. XXIV. â€” Theor. â€” If on BC the hypothenuse of a

ri^/it-angled triangle ABC, a semicircle, BAC, be described on

the same side as the triangle, and if semicircles, ADB, AEC, ba

described on the legs, falling without the triangle, the lunes or

crescents ADB, AEC, bounded by the arcs of the semicircles^

are together equal to the right-angled triangle ABC.

BOOK v.]

EUCLID AND LEGENDRE.

139

For (V, 14 and lY. 9) the semicircle ADB is to the semicircle

BAG as the square of AB to the square of BC, and AEC to

BAG as the square of AC to the square of BC ; whence (IV. 1)

the two semicircles ADB, AEG taken

+ ADC=ABDC, and AEF+AFG =

AEFG ; wherefore the squares are to

one another in the duplicate ratio of

their sides.

For like reason, the, triangles AIID,

AFC are to one another in the duplicate

ratio that PH is to DF, or that AD is to AC ; hence, similar

triangles are one to another in the duplicate ratio of their alti-

tudes or bases ; since the segments AHD, AFC have the same

bases and altitudes as the triangles AHD and AFC, and being

segments of quadrants, are similar ;

hence, they are to one another in the du-

plicate ratio of their bases and altitudes ;

therefore, by composition, AED + seg.

AHD : ABC + seg. AFC in the duplicate

ratio of their homologous sides, or simi-

lar polygons are to one another in the

duplicate ratio of their homologous sides,

and quadrants of circles are one to an-

other in the duplicate ratio of their radii ; hence, semicircles

are one to another in the duplicate ratio of their diameters, and.

circles are one to another in the duplicate ratio of their diame-

ters. Wherefore, similar surfaces are, etc.

Cor. 1. Therefore, universally, if three lines be proportionals,

the first is to the third as any plane figure upon the first to a

similar and similarly described figure upon the second.

Cor. 2. Because all squares are similar figures, and all circles

are similar figures, the ratio of any two squares to one another

is the same as the duplicate ratio of their sides ; and the ratio

of any two circles to one another is the same as the duplicate

ratio of their diameters ; hence, any two similar plane figures

are to one another as the squares or circles (IV. 7) described on

their homolocrous sides.

Cor. 3. Because the sides of similar plane figures are propor-

tional, therefore (IV. 8) their perimeters or peripheries are

propoi'tional to the homologous sides ; hence, the perimeters of

128 THE ELKMENTS OF [bOOK V.

similar polyo^ons are proportional to their apotliems ; the cir-

cumfei-enees of circles are proportional to their diameters.

Cor. 4. Hence plane figures which are similar, to the same

figure are similar to one another (I. ax. l).

Cor. 5. If four straight lines be proportionals, the similar

plane figures, similarly desci-ibed upon them, are also propor-

tionals; and, conversely., if the similar plane figures, similarly-

described upon four straight lines, be proportionals, those lines

are proportionals.

Cor. 6. Similar polygons inscribed in circles are to one an-

other as the squares of the diameters.

Cor. 1. When there are three parallelograms, AC, CH, CF,

the first, AC (IV. def 10), has to the third, CF, the ratio which

is compounded of the ratio of the first, AC, to the second, CIT,

and of the ratio of CH to the third, CF; but AC is to CH as

their bases ; and CH is to CF as their bases ; therefore AC has

to CF the ratio which is compounded of ratios that are the

same with the ratios of the sides.

Scho. 3. Dr. Simson remarks in his Note on this corollaiy,

that "nothing is usually reckoned more difficult in the elements

of geometry by learners, than the doctrine of compound ratio."

This distinguished geometer, however, has both freed the text

of Euclid from the errors introduced by Theon or others, and

has explained the subject in such a manner as to remove the

difficulties that were formeily felt. According to liim, " every

proposition in which compound ratio is made use of, may with-

out it be both enunciated and demonstrated ;" and " the use of

compound ratio consists wholly in this, that by means of it, cir-

cumlocutions may be avoided, and thereby propositions may-

be more briefly either enunciated or demonstrated, or both

may be done. For instance, if this corollary were to be enun-

ciated, without mentioning compound ratio, it might be done

as follows: If two pai-allelograms be equiangular, and if as a

side of the first to a side of the second, so any assumed straight

line be made to a second straight line ; and as the other side

of the first to the other side of the second, so the second

straight line be made to a third ; the first parallelogram is to

the second as the first straight line to the third ; and the

demonstration would be exactly the same as we now have it.

BOOK v.] EUCLTD AND LEGENDKE. 129

But the ancient cceometers, when they observed this ennncia-

ti(ni could l)e made nhortev, l)y giving a name to the ratio

which the first straiglit line lias to the last, by which name the

intermediate ratios miglA likewise be signified, of the first to

the second, and of the second to the third, and so on, if there

â– were more of them, they called this ratio of the fiist to the last,

the ratio compounded of the ratios of the first to the second,

and of the second to the third straight line; that is, in the

present example, of the ratios which are the same with the

ratios of the sides."

Scho. 4. The seventh corollary will be illustrated by the fol-

lowing proposition, which exhibits the subject in a different,

and, in some respects, a preferable light :

Triangles which have an angle of the one equal to an angle

of the othei\ are proportional to the rectangles covitcined by

the sides about those angles ; and (2) equiangular parallelO'

grams are proportional to the rectangles co7itained by their ad-

jacent sides.

1. Let ABC, DBE be two triangles, having the angles ABC,

DBE equal; the first triangle is to the second as AB.BC is to

DB.BE.

Let the triangles be placed with their equal angles coinciding,

and join CD. Then (V. 1) AB is to DB as the triangle ABC

to DBC. But (V. 1, cor. 2) AB : DB :: AB.BC :DB.BC;

theiefore (IV. 7) AB.BC is to DB.BC as the triangle ABC to

DBC. Li the same maimer it would

be shown that DB.BC is to DB.BE as

the triangle DBC to DBE ; and, there-

fore, ex ceq^io^ AB.BC is to DB.BE as

the triangle ABC to DBE.

2. If parallels to BC through A and

D, and to AB through C and E were

drawn, parallelograms would be formed which would be re-

spectively double of the triangles ABC and DBE, and which

(IV. 9) would have the same ratio as the triangles; that is, the

ratio of AB.BC to DB.BE; and this proves the second part of

the proposition.

Comparing this proposition and the corollary, we see that the

ratio which is compounded of ihe ratio of the sides, is the same

9

130

THE ELEMENTS OF

[book V.

as the ratio of their rectangles, or the same (T. 23, cor. 5) as the

ratio of their products, if they he e:q)ressed in numbers. This

conclusion might also be derived from the proof given in the

text. For (const.) DC : CE : : CG : K ; whence (V. 10, cor.)

K.DC = CE.CG. But it was proved that BC : K : : AC : CF;

or (V. 1) BC.DC : K.DC : : AC : CF; or BC.DC : CE.CG ::

AC : CF; because K.DC=CE.CG.

The twelfth proposition of this book is evidently a case of

this proposition; and the fourteenth is also easily derived

from it.

Pijop. XV. â€” TriEOPw â€” The parallelograms ahout the diago-

nal of any parallelogram are similar to the whole, and to one

another.

Let ABCD be a parallelogram, and EG, HK the parallelo-

grams about the diagonal AC ; the parallelograms EG, HK are

similar to the whole parallelogram, and to one another.

Because DC, GF are parallels, the angles ADC, AGF are

(I. 16) equal. For the same reason, be-

cause BC, EF are parallels, the angles

ABC, AEF are equal ; and (T. 15, cor. 1)

each of the angles BCD, EFG is equal

to the opposite angle DAB, and there-

fore they are equal to one another ;

Avherefore, in the parallelograms, the

angle ABC is equal to AEF, and BAC common to the two tri-

angles BAC, EAF ; therefore (V. 3, cor.) as AB ; BC : : AE :

EF. And (IV. 5), because the opposite sides of parallelograms

are equal to one another, AB : AD :: AE : AG; and DC :

BC : : GF : EF; and also CD : DA : : FG : GA. Therefore

the sides of the parallelograms BD, EG about the equal angles

are proportionals; the parallelograms are, therefore (V. def l),

similar to one another. In the same manner it would be shown

that the parallelogram BD is similar to HK. Therefore each

of the parallelograms EG, HK is similar to BD. But (V. 14,

cor.) rectilineal figures which are similar to the same figure, are

similar to one another; therefore the parallelogram EG is simi-

lar to HK. Wherefore, etc.

Scho. Hence, GF : FE : : FH : FK. Therefore the sides of

BOOK v.]

EUCLID AND LEGENDRE.

131

the paralleloojrains GK and EH, about the equal angles at F,

are reciprocally proportional ; and (V. 12) these parallelograms

are equivalent ; a conclusion which agrees with the eighth

corollary to the fifteenth proposition of the first book.

Prop. XVI. â€” Prob. â€” To describe a rectllmeal figure which

shall be similar to one given rectilineal figure., and equivalent

to one another.

Let ABC and D be given rectilineal figures. It is required

to describe a figure similar to ABC, and equal to D.

Upon the straight line BC describe (II. 5, scho.) the parallelo-

gram BE equivalent to ABC ; also upon CE describe the parallel-

ogram CM equivalent to D, having the angle FCE equal to

CBL. Therefore (I. 16 and 10) BC and CF are in a straight

line, as also LE and EM. Between BC and CF find (V. 11) a

mean proportional GH, and ^

on it describe (V. 13) the

figure GHK similar, and

similarly situated, to ABC;

GHK is the figure required.

Because BC : GH : : GH :

CF, and if three straight

lines be proportionals, as the

first is to the third, so is (V.

14, cor. 2) the figure upon the first to the similar and similarly

described figure upon the second ; therefore,

as BC to CF, so is ABC to KGH ; but (V. 1)

as BC to CF, so is BE to EF ;

therefore (IV. 7) as ABC is to KGH, so is BE to EF. But

(const.) ABC is equivalent to BE ; therefore KGH is equivalent

(IV. ax. 4) to EF; and (const.) EF is equivalent to D; where-

fore, also, KGH is equivalent to D; and it is similar to ABC.

Therefore the rectilineal figure KGH has been described, simi-

lar to ABC and equivalent to D ; which was to be done.

Prop. XVII. â€” Theor. â€” If two similar parallelograms have

a comm,on angle, and be similarly situated, they are about the

same diagonal.

Let ABCD, AGEF be two similar parallelograms having a

132

THE ELEMENTS OF

[book V.

C

E

F

B

common angle CAB, tlicy will be about the same diagonal

AD. Similar parallelograms have

their sides about equal angles pro-

portional (V. def. l). Draw the

diagonals EG and CB ; hence, AB :

AG : : AC : AE ; therefore EG ia

parallel to CB (V. 3), and the angles

AEG, ACB are equal (I. 16); like-

wise the angles EGA, CBA. The

triangles E AG, FG A are equal (I. 15, cor. 5); likewise the tri-

angles CAB, DBA ; therefore AF is equal to EG, and AD is

equal to CB ; but AF is the diagonal also of AGEF, and is in

the same straight line with AD, the diagonal of ABCD.

Wherefore, if two similar parallelograms, etc.

Cor. Hence, equiangular parallelograms have to one another

the ratio which is compounded of the ratio of their sides;

hence, triangles which have one angle of the one equal, or sup-

plemental, to one angle of the other, have to one another the

ratio which is compounded of the ratio of the sides containing

those angles.

Pkop. XVIII. â€” ^Theok. â€” Of all the parallelof/rmns that can

he inscribed in any triangle^ that which is described on the half

of one of the sides as base is the greatest.

Let ABC be a triangle, having BC, one of its sides, bisected

in D; draw (I. 18) DE parallel to BA, and EF to BC ; let also

G be any other point in BC, and

describe the parallelogram GK;

FD is greater than KG.

If G be in DC, through C draw

CL parallel to BA, and produce

FE, KH, Gil as in the figure.

Then (I. 15, cor. 8) the comj>le-

ments LTI and IID are equivalent ; and since the bases CD,

DB are equal, the parallelograms ND, DK (I. 15, cor. 5) are

equivalent. To LH add ND, and to HD add DK ; then (I.

ax. 2) the gnomort MND is equivalent to the parallelogram

KG. But (L ax. 9) DL is greater than MXD ; and therefore

BOOK V.J EUCLID AND LFGENDEE. 133

FD, which (T. 15, cor. 5) is equal to DL, is greater than KG,

which is equivalent to the gnomon JNIND. -

If G were in BD, since BD is equal to DC, AE is equal (V.

2) to EC, and AF to FB; and by drawing through A a jiaral-

lel to BC, meeting DE produced, it would he proved in the

same manner that FD is greater than the inscribed ])aralk'lo-

gram applied to BG. Therefore, of all the parallelogi-ams, etc.

Cor. Since (V. lo) all parallelograms having one angle coin-

ciding with BCL, and their diagonals with CA, arc similar, it

follows from this proposition that if, on the segments of a given

straight line, BC, two parallelograms of the same altitude be

described, one of them, DL, similar to a given parallelogram,

the other, DF, will be the greatest possible when the segments

of the line are equal.

Scho. The parallelogram FD exceeds KG by the parallelo-

gram OM similar to DL or DF, and described on OH, which

is equal to DG, the difference of the bases BD and BG. Hence

we can describe parallelograms equivalent and similar to given

rectilineal figures.

The enunciation of this proposition here given is much more

simple and intelligible than that of Euclid, and the proof is

considerably shortened, Euclid's enunciation, as given by Dr.

Simson, is as follows: "Of all pai'allelograms applied to the

same straight line, and deficient by parallelograms, similar and

similarly situated to that which is described upon the half of

the line ; that which is applied to the half, and is similar to its

defect, is the greatest." It may be remaiked, that this piopo-

sition, in its simplest case, is the same as the second corollary

to the fifth proposition of the second book.

Prop. XIX. â€” ffHEOR. â€” In equal circles^ or in the same cir-

cle^ angles, whether at the centers or circumferences^ have the

same ratio as the arcs on which they stand have to one another^

so also have the sectors.

Let, in the equal circles ALB, DNE, the angles LGK, KGC,

CGB, DHN, NHM, and IMITF be at the centers, and the angles

BAG and EDF be at the circumferences, then will those angles

have to each other the same ratio as the arcs KL, KC, CB, DN,

NM, MF, and FE have to one another.

134

THE ELEMENTS OF

[book V.

Since (T. def. 19) all angles at the center of a circle are

measured by the arcs intercepted by the sides of the angles,

and all ancjles at the circum-

ference are subtended by the

arcs intercepted by the sides

of the angles, and (III. 16)

equal angles will have equal

arcs whether they be at the

center or the circumference;

hence, the same ratio which

the arcs have to each other, will the angles also have to one

another â€” that is, when the arcs be greater, the angles will be

greater; less, less ; and equal, equal.

And since the sectors are contained (III. def V) by the sides

of the anirles and the arcs, the same ratio between the sectors

will evidently exist as there is between the arcs. Wherefore,

in equal circles, etc.

Cor. Hence, conversely^ arcs of the same or equal circles will

have the same ratio as the angles or sectors which they measure

or subtend â€” when greater, greater ; less, less ; or equal, equal.

Pkop. XX. â€” ^Peob. â€” Tlie area of a regular inscribed poly '

gon^ and that of a regular circumscribed one of the same nvirv-

her of sides being given y to find the areas of the regidar in-

scribed and circumscribed polygons having double the number

of sides.

].et A be the center of the circle, BC a side of the inscribed

polygon, and DE parallel to BC, a side of the circumscribed

one. Draw the perpendicular AFG, and the tangents BH,

CK, and join BG; then BG will be a side

of the inscribed polygon of double the

number of sides; and (111.^6, cor. l) IIK is

a side of the similar cii-cumscribed one.

/ \\ // \ Now, as a like construction would be

/ x/^ \ Tinule at each of the remaining angles of

MAN ^^Ijp iwlygon, it will be sufficient to con-

sider the i)art here represented, as the tri-

angles connected with it are evidently to each other as the poly-

gons of which they are parts. For the sake of brevity, then.

n

G K

BOOK v.] EUCLID AND LEGENDKS. 135

let P denote the polygon whose side is BC, and P' that whose

side is DE; and, in like manner, let Q and Q' represent those

whose sides are BG and HK ; P and P/ therefore, are given ;

Q and Q' required.

Now (V. ]) the triangles ABF, ABG are proportional to

their bases AF", AG, as they are also to the polygons P, Q ;

therefore AF : AG : : P : Q. The triangles ABG, ADG are

likewise as their bases AB, AD, or (V. 3) as AF, AG; and

they are also as the polygons Q and P' ; therefore AF : AG : :

Q : P' ; wherefore (IV. 7)" P : Q : : Q : P' ; so that Q is a mean

proportional between the given polygons P, P'; and, re))re-

senting them by numbers, we have Q-=PP'', so that the area

of Q will be comjyuted by -tnaltlphjing P by P', and extracting

the square root of the product.

Again : because AH bisects the angle GAD, and because

the triangles AHD, AHG are as their bases, Me have Gil : HD

: : AG : AD, or AF : AB : : AHG : AHD. But we have

already seen that AF : AG :: P : Q; and therefore (IV. 7)

AHG : AHD : : P : Q. Hence (IV. 11) AUG : ADG : : P :

P-|-Q; whence, by doubling the antece-

dents, 2 AHG : ADG : : 2P : P+Q. But

D IT P K" "R

it is evident, that whatever part the tri-

angle ADG is of P^, the same part of the

polyg )n Q' is the triangle AHK, which is

double of the trianojle AHG, Hence the

last analogy becomes Q' : P^ : : 2P : P+Q.

Now (IV. 2, cor. 1) the product of the ex-

tremes is equal to the product of the means; and therefore Q'

will be computed by dividing twice the product o/'P and V by

P+Q; and the mode of finding Q has been pointed out

already.

Prop. XXI. â€” ^Theor. â€” Of regular polygons which have

eqiial perimeters., that which has the greater mimber of sides is

the greater.

Let AB be half the sides of the polygon which has the less

number of sides, and BC a perpendicular to it, which will evi-

dently pass through the center of its inscribed or circumscribed

circle; let C be that center, and join AC. Then, ACB will be

136

THE ELEMENTS OF

[book V.

the angle at tlie center subtended by the half side AB. ]\I;ike

BCD equal to the angle subtended at the center of the other

polygon by half its side, and from C as center, with CD as ra-

dius, describe an arc cutting AC in E, and CB i)roduced in F.

Then, it is plain, that the angle ACB i'* to four right angles as

AB to the common peiimeter; and four right angles are to

DCB, as the common pei-imeter to the half of

a side of the other polygon, which, for brevi-

ty, call S; then, ex mquo^ the angle ACB is

to DC]} as AB to S. But (V. 19) the angle

ACB is to DCB as the sector ECF to the

sector DCF; and consequently (IV. 1) the

sector ECF is to DCF as AB to S, and, by

division, the sector ECD is to DCF as AB â€”

S to S. Now the triangle ACD is greater

than the sector CED, and DCB is less than DCF. But (V. 1)

these triangles are as their bases AD, DB ; therefore AD has to

DB a greater i-atio than AB â€” S to S. Hence AB, the sum of

the first and second, has to DB, the second, a greater ratio than

AB, the sum of the third and fourth, has to S, the fourth ; and

therefore (IV. 2, cor. 5) 8 is greater than DB. Let then BG

be equal to S, and draw GH parallel to DC, meeting FC pro-

duced in H. Then, since the angles GHB, DCB are equal, BH

is the perpendicular drawn from the center of the polygon hav-

ing the greater number of sides to one of the sides ; and since

this is greater than BC, the like perpendicular in the other

polygon, while the perimeters are equal, it Ibllows that the area

of that which has the gi-eater number of sides is greater than

that of the other.

Prop. XXIT. â€” Theor. â€” If the diameter of a circle 1 e divided

into any two parts, AB, BC, and if semicircles, ADBjBECy be

described on opposite sides of these, the circle is divided by

tJieir arcs into ttoo figures, GDE, FED, the boundary of each <f

tchich is equal to the circumference q/'FG ; and which are such

that AC : BC : : FG : FED, and AC : AB :: Â¥G : GDE.

For (V. 14, cor. 3) the circumferences of circles, and conse-

quently the halves of their circumferences, are to one another as

their diameters ; thereiore AB is to AC as the arc ADB to

BOOK v.] EUCLID AND LEGENDRE. 137

AFC, and BC is to AC as the arc BEC to AFC. Hence (IV.

1) AC is to AC as the compound arc

ADEC to AFC; therefore ADEC is p

equal to half the circumference ;and the

entire boundaries of the fiirures GDE,

FED are each equal to the circumfer-

ence of FG.

Again (V. 14, cor, 2): circles, and

consequently semicircles, are to one an-

other as the squares of their diameters ;

therefore AC' is to AB^ as the semicircle

AFC to the semicircle ADB. Hence, since (H. 4) AC' = ABÂ»

-f2AB.BC + BC^ we find by conversion that AC is to 2AB.

BC + BC% as the semicircle AFC to the remaining space

BDAFC; whence, by inversion, 2AB.BC + BC' is to AC' as

BDAFC to the semicircle AFC. But BC' is to AC= as the

Bemicircle BEC to the semicircle AFC; and therefore (IV. 1)

2AB.BC + 2BC' is to AC as the compound figure FDE to the

semicircle AFC. But (II. 3) 2AB.BC+2BC = 2AC.BC, and

(V. 1) 2AC.BC : AC^ : : BC : ^AC. Hence the preceding

analogy becomes BC to ^AC, as FED to the semicircle, or by

doubling the consequents, and by inversion, AC to BC, as FG

to FED ; and it would be proved, in the same manner, that

AC : AB : : FG : GDE.

Cor. Hence we can solve the curious problem, in which it is

required to divide a circle into any proposed number of parts,

equal in area and boundary; as it is only necessary to divide

the diameter into the proposed number of equal parts, and to

desci'ibe semicircles on opposite sides. Then, whatever part

AB is of AC, the same part is AEG of the circle. Their bound-

aries are also equal, the boundary of each being equal to the

circumference of the circle.

Scho. Another solution would be obtained, if the circumfer-

ence were divided into the proposed number of equal parts, and

radii drawn to the points of division. This division, hoM'ever,

can be eftected only in some particular cases by means of ele-

mentary geometry.

Pkop. XXIII. â€” ProbÂ» â€” To divide a given circle ABC into

138

THE ELEMENTS OF

[book V.

any proposed number of equal parts by means of concentric

circles.

Divide the radius AD into the proposed number of equal

parts, suppose three, in the points E, F, and through these

points draw perpendiculars to AD, meeting a semicircle de-

scribed on it as diameter in G, H;

from D as center, at the distances DG,

DH, describe the circles GL, HK ; their

circumferences divide the circle into

equal parts.

Join All, DH. Then (V. 17, cor. 2)

AD, DH, DF being continual propor-

tionals, AD is to DF as a square de-

scribed on AD is to one described on

DH. But (V. 14, cor. 2) circles are proportional to the squares

of their diameters, and consequently to the squares of their

radii. Hence (IV. 7) AD is to FD as the circle ABC to the

circle HK ; and therefore, since FD is a third of AD, HK is a

third of ABC. It would be proved in a similar manner that

AD is to ED as ABC to GL. But ED is two thirds of AD,

and therefore GL is two thirds of ABC ; wherefore the space

between the circumferences of GL, HK is one third of ABC,

as is also the remaining space between the circumferences of

ABC and GL.

Cor. Hence it is plain that the area of any annulus^ or nng,

between the circumferences of two concentric circles, such as

that between the circumferences of ABC and GL is to the cir-

cle ABC as the difference of the squares of the radii DM, DL

to the square of DM ; or (II. 5, cor. 1, and HI. 20) as the rect-

angle AL.LM, or the square of the perpendicular LB to the

square of DM; and it therefore follows (V. 14) that the ring is

equivalent to a circle described with a radius equal to LB.

Prop. XXIV. â€” Theor. â€” If on BC the hypothenuse of a

ri^/it-angled triangle ABC, a semicircle, BAC, be described on

the same side as the triangle, and if semicircles, ADB, AEC, ba

described on the legs, falling without the triangle, the lunes or

crescents ADB, AEC, bounded by the arcs of the semicircles^

are together equal to the right-angled triangle ABC.

BOOK v.]

EUCLID AND LEGENDRE.

139

For (V, 14 and lY. 9) the semicircle ADB is to the semicircle

BAG as the square of AB to the square of BC, and AEC to

BAG as the square of AC to the square of BC ; whence (IV. 1)

the two semicircles ADB, AEG taken

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 11 of 21)