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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 LibraryLawrence S. (Lawrence Sluter) BensonGeometry : 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
■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. — TriEOPw — 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. — Prob. — 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. — Theor. — 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. — ^Theok. — 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. — ffHEOR. — 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 — 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 — when greater, greater ; less, less ; or equal, equal.

Pkop. XX. — ^Peob. — 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. — ^Theor. — 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 —
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 — 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. — Theor. — 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 :: ¥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' = AB»
-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. — Prob» — 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. — Theor. — 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


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Online LibraryLawrence S. (Lawrence Sluter) BensonGeometry : 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)