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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 4 of 21)
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therefore the angle ADB is equal to the angle ABD (I. 1, cor. 1).
But (I. 20) the exterior angle ACB is equivalent to the sum of
the two remote interior angles CDB and DBC. And CDB is
equal to DBA (I. 1, cor. 1), but ABC is less than ABD (I. ax.
9) ; therefore ACB is greater than CBA.

The proof can also be given by laying oflf on the greater side





38 THE ELEMENTS OF [bOOK I.

AB, tlie less side AC ; joining the vertex of the opposite angle
with the point where AC terminates on AB ; and the demon-
stration conducted similarly to the preceding. Again : by
cutting oflf on AC a part equal to CB, hisect the angle BCA,
and join the extremity of the part on AC equal to CB with the
foot of the line bisecting the angle BCA ; this proof is given by
means of (I. 3 and 20).

Conversely: when the angle ACB is greater than the angle
ABC, the side AB is greater than AC. At the vertex of BCA

on AC make an angle ACD equal
to the angle ABC (I. 13). Now,
the angle ACD, the equal to the
angle ABC, is subtended by AD,
~^ ^ ^ and the angle ACB is subtended by

AB, but AB is greater than AD ; hence the greater angle is
subtended by the greater line ; therefore, in the triangle ACB,
the greater angle ACB is subtended by a greater side than the
less angle ABC. And (L def. 19) if the angles ACD and ABC
be subtended by arcs, the arc subtending ACB is greater than
the arc subtending ABC ; but the side AB is the chord of the
arc subtendins: ACB, and AC is the chord of the arc subtending
ABC ; therefore AB is greater than AC. Wherefore, if two
sides of a triangle, etc.

Cor. Hence any two sides of a triangle are together greater
than the remaining side.

Scho. The truth of this corollary is so manifest, that it is
given as a corollary to avoid increasing the number of axioms.
Archimedes defined the straight line the shortest distance be-
tween two points ; hence two straight lines connecting three
points not in the same direction, are together greater than one
straight line connecting any two of those jDoiuts.

Prop. XXII, — Tiieor. — If two triangles have tico sides of
the one equal to two sides of the other, each to each, hut the
angles contained hg those sides loiequal, the base or remaining
side of the one ichich has the greater angle is greater than the
base or reraaining side of the other.

Let DEG and DEF be two triangles which have the sides
DE common, the sides DG and DF equal, but the angle EDG




HOOK I.] EUCLID AND LEGENDEE. 39

greater tlian the angle EDF ; the side EG is also greater than
the side EF.

In the triang-les DEF and DEG Ave have
(hyp.) DE commoii, DG equal to DF, and
the an'He EDG crreater than the anHe
EDF. Now, because DG and DF are equal,
the angles DGF and DFG are equal (I. 1,
cor. 1) ; but the angle DGF is greater than
the angle EGF (I. ax. 9) ; therefore the angle
DFG is greater than the angle EGF ; and much more is the
angle EFG greater than the angle EGF. Then (I. 21), EG op-
posite EFG is greater than EF opposite EGF. Wherefore, if
two triangles have, etc.

There are other cases of this proposition ; if a line equal to
DE, the less side, be drawn through D, making with DF, on the
same side of it with DE, an angle equal to EDG, the extremity of
that line might fall on FE produced, or above, or helo^o it. Or
the ano-le DFE could be in the triangle DEG, or on the other
side of DE. And a very easy proof can be given by bisecting
the angle FDG by a straight line cutting EG in a point, which
call K, and joining DK and KF, for KG would be equal to
KF (I. 3) ; adding EK, we would have EG equivalent to EK
and KF; therefore (I. 21, cor.) EG greater than EF.

Cor. Hence, conversely, if two triangles have two sides of
the one equal to two sides of the other, each to each, but their
bases unequal, the angles contained by the respectively equal
sides of those triangles arc also unequal, the greater angle
being in the triangle which has the greater base. For with a
radius, DG, or its ecpial, DF, describe a circle (I. post. 3) from a
center, D, and draw on the same side of DE with the angle
EDF a line equal to EG from the other extremity of DE to the
circumference, then in the triangles DEG and DEF we have
(const.) DG equal to DF (I. def 16). But (hyp.) the side EF
is less than EG ; hence the angle EDF is less than the angle
EDG (I. ax. 9).

Pkop. XXin. — Peob. — To describe a parallelogram upon a
given straight line.

Let DB be a given straight line ; it is required to describe a



40



THE ELEMENTS OF



[book I.



paralleloGjram. U])Ou it. From a point, D, on DB draw DF (I.



F



H




B



post. 1), then from the point F on
DF draw FL parallel to DB (I. 18) ;
and if through B, a point on DB, a
parallel to DF (I. 18), he drawn,
DBFG (I. def. 15) is a parallelogram.
If from D a perpendicular, DII (I. 7), be drawn, then the
angle HDB is a riglit angle (I. def 10) ; and from H a parallel
to DB as HL be drawn (I. 18), and from B a parallel to HD
as LB be drawn (I. 18), then DBIIL (I. def 15) is a rectangle.
And if from D as a center, and a radius DB (I. post. 3), an arc
be described intercepting DH, or DH produced, and a rectangle
be described from that point where DH is intercepted, that
rectangle will be a square (I. def 15).

Co?: 1. Hence (const, and I. 15, cor. l) a square has all its
sides equal, and (I. 15, cor. l) all its angles are right angles.

Cor. 2. Hence the squares described on equal straight lines
(I. 15, cor. 4) are equal.

Cor. 3. If two squares be equal, their sides are equal (I.
ax. 1).

Cor. 4. If AB and AD, two adjacent sides of a rectangle
BD, be divided into parts which are all equal, straight lines
drawn through the points of section, parallel to the sides, divide
the rectangle into squares which are all equal, and the number
A B of which is equal to the product of the

number of parts in AB, one of the sides,
multiplied by the number of parts in
AD, the other. For (const.) these



figures are all parallelograms; and (I.
c def. 15 and const.) the sides being equal,
and the angles being (I. 16, part 2) equal to A, and there-
fore right angles, hence (I. 15, cor. 4) they are all squares. Of
these squares, also, there are evidently as many columns as there
are parts in AB ; while in each column there are as many squares
as there are parts in AD. The number of such squares con-
tained in a figure is called, in the language of mensuration, the
area of that figure.

Cor. 5. Hence, since any parallelogram is equivalent (I. 15,
cor. 5) to a rectangle on the same base and between the same



BOOK I.] EUCLID AND LEGENDKE. 41

parallels, it follows that the area of any 2^'^^'>'(Melogram is
equivale7it to the product of its base and its "perpendicidar
height / and the area of a square is computed by multiplying
a side by itself

Cor. 6. Hence, also (I. 15, cor. 4), the area of a triangle is
computed by m.idtiplying any of its sides by the perpendicidar
draion to that side from the opposite angle., and taking half
the product ; and the area of a trapezium is found by multi-
P'lying either diagonal by the sum of the perpendicidar s draion
to it from the ayigles which it subtends, and taking half the
product. When, in consequence of one of the angles being re-
entrant, the perpendiculars lie on the same side of the diagonal,
the difference of the perpendiculars must evidently be used in-
stead of their sum.

Cor. 7. Every polygon may be divided into triangles or tra-
peziums by drawing diagonals ; and therefore the area of any
polygon whatever can be computed by finding the areas of
those component figures by the last corollary, and adding them
together.

Scho. This corollary and the two foregoing contain the ele-
mentary principles of the mensuration of rectilineal figures, and
they form a connection between arithmetic or algebra and ge-
ometry. They also explain the origin of the expressions, " the
square of a number," " the rectangle of two numbers," and " the
product of two lines."

Prop. XXIV. — Theor. — If parallelograms be described on
two sides of any triangle, and their sides which are parallel to the
sides of the triangle be produced until they meet, the sum of
the parallelograms will be equivalent to the parallelogram de-
scribed on the base of the triangle having its adjacent sides to
the base parallel to the straight line joining the vertex of the
triangle with the point of intersection of the sides of the other
parallelograms produced, and terminated by the latter sides or
those sides produced.

Let BAG be a triangle ; the parallelograms MB AD and CEFA,
described on the two sides BA and CA, respectively, are to-
gether equivalent to the parallelogram BCHK, described on
the base BC ; the parallelograms MBAD, CEFA, and BCIIK



43



THE ELEMENTS OF



[book I.




beino; described ao;reeablv to the
proposition.

Describe on BA and CA (L
23) tlie parallelograms MB AD
and CEFA ; and produce the
M B L c E titles MD and EF until they

meet in G ; draw GA, and produce it to L on the base BC (I.
post. 2) ; describe the parallelogram BCIIK (I. 23) on BC.
Then, since (const, and I. 15, cor. 1) BH and CK are parallel
and equal to AG, they are parallel and equal to one another (I.
ax. 1) ; also (1. 15, cor 3) HK is parallel and equal to BC ; hence (I.
def. 15) BCHK is a parallelogram, and BLIIW and LCWK are
also parallelograms. Xow, the parallelograms BLHW and
HBAG (I. 15, cor. 5) are equal, and for similar reason the parallel-
ograms HBAG and MBDA ; hence (I. ax. 1) MBDA is equal to
BLHW. In similar manner, it can be shown tlnit CEFA is
equiA^alent to LCKW; therefore the whole parallelogram BCHK
(I, axs. 2 and 10) is equivalent to the sum of the two parallelo-
grams MB AD and CEFA. Wherefore, if on any two sides of
a triangle, etc.

Cor. 1. A paiticular case of this proposition is, lohen the tri-
angle is right-angled., then the sqiuires described on the legs —
that is, the sides containing the right angle, are together equiva-
lent to the square on the liypotlieniise — that is, the side opposite
the right angle.

Let ABC be a rio-ht-anoled
triangle, having the right angle
BAC ; the square described on
tlie hypothenuse BC is equiva-
lent to the sum of the squares
on BA and AC. On BC, BA,
Q and AC (L 23) desciibe the
squares BCHK, BADE, and
ACFG ; through A draw AL
parallel to BH (L 18), and
draw AH and DC (L post. 1).
Tlien, because the angles
BAC and BAE are both right
angles (L def 10), the two




BOOK I.]



EUCLID AND LEGENDEE.



43



straight lines CA and AE are in the same straight line (I, 10).
For like reason, BA and AF are in the same straight line.
Again : because the angle IIBC is equal to the angle DBA (I,
ax. 1), if the angle ABC be added to each, we have (I. ax. 2)
HBA equal to DBC ; and because AB is equal to DB (const.),
and BIT equal to BC, therefore (I. 3, case 1) the triangle ABH
is equal to the triangle DBC. But (I. 15, cor. 4) the parallelo-
gram BPIIL is double the ti-iangle ABH ; for like reason the
square BADE is double the triangle DBC ; hence (I. ax. 6)
BPIIL is equivalent to BADE. In like manner, PCLK can be
shown equivalent to ACFE. Xow (I. ax. 10), BCHK is equiva-
lent to BPHL and PCLK together; hence (L ax. 1) BCHK is
equivalent to BADE and ACFG together. Wherefore, if the
triangle is rio-ht-angled, etc.




OTHERWISE,

Let the squares on AB and
BC fall on the same side of BC.
Describe the square BAED on
the side BA (L 23), and th.e
square BCMN" on the side BC
(L 23), and produce AE to F
(I. post. 2) ; then tlirough D draw
PF parallel to BC (I. 18).

Because AF is parallel to BD
(L def. 14, and 15, cor. 1), BC is
equal to DF, and BA is equal to '^^
DE, and the angles BAC and DEF are both right angles (1.
def 10), and equal (I. ax, 11); therefore the triangle BAC is
equal to the triangle DEF (I. 3) ; and because BCED is com-
mon to the square BAED and the parallelogram (I. def 14)
BCDF, and the triangles BAC and DEF equal, the square
BAED is equivalent to the parallelogram BCDF. And PF
being parallel to BC (const.), the parallelograms (I. def 14)
BCDF and BCPL have a common base, BC, and equal altitudes;
hence (I. 15, cor. 5) they are equivalent, and (I. ax, 1) the
square BAED is equivalent to the parallelogram BCPL. From
A draw (I. 18) AK parallel to BM, and produce DE (I. post.
2) to BM; then AK and BM being parallel (const.), ED and



44



THE ELEMENTS OF



[book I.



BA, being opposite sides of the same square, are also parallel
(I. def. 15) ; hence MR is equal to BA, and equal also to DE
(I. ax. 1). But BM is equal to BC (const.) ; therefore the par-
allelogram BAjMR is equal to the parallelogram BCDF ; and
BAMR having the same base and equal altitude with the par-
allelogram BGMK, is equivalent to it (I. 15, cor. 5); hence
BGMK is equivalent to BCDF, equivalent to BCPL, aud
equivalent to the square BADE (I. ax. 1).

Or, the square described upon BA is equivalent to the rect-
angle of the hypothenuse B C and the part BG of the hypjoth-
enuse nearest to BA intercepted by the p)erpe7idicular drawn
from the vertex of the right angle to the liypothenuse. In a
similar manner, it can be shown that the square described on
A G is equivalent to the rectangle of the hypothenuse B (7, and
the remaining part G G of the hypothenuse intercepted by the
same p)erpendicxdar. But the two rectangles are equivalent to
the square of the hyi^othenuse (I. ax. 10) ; hence the tico squares
described on the sides AB and AG (I. ax. 1) are equivalent to
the square described on the hypothenuse.

E F And if we make the tiiangle

an isosceles rio;ht-an(2;led tri-
angle as ABC, the square de-
sciibed on AB will contain
four equal triangles, ACB,BCF,
FCE, and ECA, while each of
the squares described on AC
and CB will contain two such triangles, and both together will
be equivalent to the four equal triangles, or equivalent to the
square on AB. The demonstration is very simple, and it would
be well for the pupil to undertake it.

Hence, conversely, if the square described upon one side of a
triangle be equivalent to the sum of the square described upon
the two other sides of the triangle, the angle contained by those
two sides is a right angle ; and when those two sides form two
equal squares, the triangle is a right-angled isosceles triangle.

Scho. 1. The proof of the corollary can be shown, also, either
by describing the square of the hypothenuse on the other side
of BC; and the other squares sometimes on one side and some-
times on the other; and since drawing a perpendicular from the




BOOK I.] EUCLID AND LEGEXDKE. 45

vertex of tlie riglit angle to the hypothenuse makes two ri(iht
angles, so a line can be drawn from the same vertex to the
point of bisection of the hypothenuse and make two supple-
niental avgles and two equivalent triangles, and the demonsti'a-
tion conducted by supplemental angles (I. 15, cor. V) instead of
right angles. Proportion also gives neat and easy solutions to
this corollary. (See V. 8, scho.)

Cor. 2. If two right-angled triangles have their hypothenuses
equal, and a side similarly situated in each also equal, the two
triangles are equal by the third proposition of this book ; and,
conversely, if the legs of a right-angled triangle be equal to the
legs of another right-angled triangle, each to each, their hypoth-
enuses can be in a similar manner shown equal.

Cor. 3. Hence, also, we can find a square equivalent to the
sum of more than two squares ; thus, let AB be the side of one
square, and AC, perpendicular to it, the side of another squai'e ;
join CB ; the square on CB (I, 24, cor. l)
is equivalent to the sum of the squares on
CA and AB. In like manner, if CD be
drawn perpendicular to CB, and DB be
drawn, the square on DB is equivalent to
the squares on DC, CA, and AB, and by
drawing a perpendicular to DB, a square can be found equiva-
lent to the sum of four squares ; hence a square can be found
equivalent to the sum of any number of squares.

Cur. 4. Since (CB)- o (AB)= + (CA)=, we have (AB)=o
(CB)" — (CA)-; hence a square can be found equivalent to the
difference of two squares.

Cor. 5. If a perpendicular be drawn from the vertex of the
angle A, in the triangle BAC (diagram to cor. l), to P on the
hypothenuse BC, cutting BC into two segments, BP and PC,
the difference of the squares on the sides AB and AC is equiva-
lent to the difference of the squares on the segments BP and
PC. For the square on AB is equivalent to the squares on BP
and PA, and the square on AC is equivalent to the squares on
PC and PA; therefore (AC)^ — (AB)^ =o- (PC)' — (BP)^

Cor. 6. Hence the squares on txoo sides of a triangle are to-
gether equivalent to txoice the square of half the remaining side^
and twice the square of the straight line from its point of M-




46



THE ELEMENTS OF



[book I.



section to the opposite anrjle. Suppose P in tlic triangle to te
the point of bisection of the side BC ; tlion, when AP is per-
pendicuhir to BC, we have (AB)- + (AC)= o (BP)- + (AP)'
+ (PC)' + (AP)^ .0= 2 (BP)= + 2 (AP)\ And when AP is
not perpendicular to BC, the equivalence of the squares on two
sides to the same will be shown in tlic next book.

Scho. 2. In proof of coj-. 5, the obvious principle is employed,
that the difference of two magnitudes is the same as the differ-
ence obtained after adding to each the same third magnitude.
Thus the difference of the squares on BP and PC is the same as
the difference between the sum of the squares BP and PA and
of PC and PA.



Pkop. XXy. — Theok. — Tlie side and diagonal of a square
are incommeiisuraUe to one another — that is, there is no Ivie
which is a measure of both.

Let ABCD be a square, and BD one of its diagonals ; AB,
BD are incommensurable.

Cut off DE equal to DA, and join AE. Then, since (I. 1,

cor. 1) the angle DEA is equal to the acute angle DAE, AEB

A B i^ obtuse, and therefore (I. 22, cor.) in

the triangle ABE, BE is less than AB,
or than AD ; wherefore AD is not a
measure of BD. Draw EF ])erpendicular
to BD. Then tiie angles FAE, FEA,
being the complements of the equal
angles DAE, DEA, are equal, and
therefore AF, FE are equal. But (I.
20, cor. 4) ABD is half a riglit angle ;
as is also BFE, since BEF is a right angle ; wherefore BE is
equal to FE, and therefore to AF. From FB, which is evi-
dently the diagonal of a square of which FE or EB is the side,
cut off FG equal to FE, and join GE. Then it would be shown,
as before, that BG is less than BE ; and therefore BE, the dif-
ference between the side and diagonal of tlie square .VC, is con-
tained twice in the side AB, with the remainder GB, wliich is
itself the difference between the side FE or EB, and the diag-
onal FB of another squai*e. By repeating the process, we
Bhould find, in exactly the same manner, that BG would be




BOOK I.] EUCLID A:SD LEGENDEE. 47

contained twice in BE, with a remainder, which would be the
difierence between the side and diagonal of a square described
on I3G ; and it is evident that a like process might be repeated
continually, as no excess of a diagonal above a side would
be contained in the side witliout remainder; and as this pro-
cess has no termination, theiv is no line, however small, which
will be contained without remainder in both AB and BD ; they
are, therefore, incommensurable.

Scho. Tills proposition can be illustrated by numbers. Let
10 be the side of tlie square ; then (I. 24, cor. 1) the diagonal
will be expressed by the square root of 200, or 14'142 + ; there
being no common multiple of 10 and 14-142 +, these numbers
are incommensurable with each otlier. Or, wlicn any two lines
are taken which by division and subdivision no common meas-
ure can be found which can be contained in each v.'ithout a re-
mainder, the two lines are said to be ijicommensicrable with
each other, and such lines are tlie side and diagonal of a square ;
and any two magnitudes wiiatever which have no common
unit of measure are incommensurable with one another.



E^TD or BOOK FIRST.



BOOK SECOND.*
ON" THE llECTANGLE AND SQUARE.

DEFINITIONS.

1. A rectangle is said to be contained by the two straight
lines which are about any of the right angles. For the sake of
brevity, the rectangle contained, by AB and CD is often ex-
pressed simply by AB. CD, a jooint being placed between the
letters denoting the sides of tlie rectangle ; and the square of a
line AB is often written simply AB".

2. A gno7non is the part of a parallelogram which remains
when either of the parallelograms about one of the diagonals
is taken away.

PROPOSITIONS.

Pkop. I. — TiiEOR. — If there be tioo straight lines, one of
which is divided into any number of parts, the rectangle con-
tained by the two lines is equivalent to the rectangles contained
by the undivided line, and the several j^arts of the divided line.

Let A and BC be two straight lines; and let BC be divided
into any parts in tlie points D, E ; tlie rectangle contained by
A and BC is equivalent to the rectangles contained by A and
BD, A and DE, and A and EC.

From B draAV (I. V) BG perpendicular to BC, and make it
equal to A ; through G draw (I. 1 8) GH par-
^ DEC ^^^^^ ^^ J3Q . ^^^^ through D, E, C draw DK,

EL, CH parallel to BG. Then BH, BK, DL,
and EH are evidently rectangles ; and BH is
equivalent (I. ax. 10) to BK, DL, EH. But
BH is contained by A and BC, for (H. dcf. 1)
■^ it is contained by GB and BC, and GB is

* The second and third books are arrantred very similarly to those
books in the edition of Euclid i)y Professor Thomson of the University
of Glasgow, Scotland.



G K L n



BOOK 11.^ EUCLID AND LEGENDEE. 49

equal (const.) to A; and BK is contained by A and BD, for it
is contained by GB and BD, of whicli GB is equal to A. Also
DL is contained by A and DE, because DK, that is (I. 15, con
4) BG, is equal to A ; and in like manner it is shown that EH
is contained by A and EC. Therefore the rectangle contained
by A and BC is eqidvalent to the several rectangles contained
by A and BD, by A and DE, and by A and EC. Wherefore,
if there be two straight lines, etc.

Prop. II. — Theoe. — If a straight line he divided into any
ttoo parts, the rectangles contained by thexchole and each of the
parts are together equivalent to the square of the whole line.

Let the straight line AB be divided into any two parts in the
point C ; the rectangles AB.AC and AB.BC are equivalent tc»
the square of AB,

Upon AB describe (I. 23) the square AE, and through C
draw (I. 18) CF parallel to AD or BE. Then
AE is equal (I. ax. 10) to the rectangles AF and
CE. But iVE is the square of AB, and AF is
the rectangle contained by BA, AC ; for (II,
def 1) it is contained by DA, AC, of which DA
is equal to AB ; and CE is contained by AB,
BC, for BE is equivalent to AB ; therefore the
rectangles under AB, AC, and AB, BC are equivalent to the-
square of AB. If, therefore, etc.

8cho. This proposition may also be demonstrated in the fol-
lowing manner :

Take a straight line D equal to AB, Then (II, 1) the rect-
angles AC.D and BCD are together equiva-
lent to AB.D. But since D is equal to AB, ^ ^ ^

the rectangle AB.D is equivalent (I. def 15)

to the square of AB, and the rectangles

AC.D and BCD are respectively equivalent °

to ACAB and BCAB ; wherefoi-e the rect-
angles ACAB and BCAB are together equivalent to the
square of AB.

In a manner similar to this, several of the following proposi-
tions may be demonstrated. Such proofs, thougli perhaps not
60 easily understood at first by the learner, are shorter than
4





so THE ELEMENTS OF [bOOK H.

those given by Euclid; aud they have tlie advantage of being
derived from those preceding them, instead of being estab-
lished by continual appeals to original principles.

Pkop. III. — Theor. — If a straight line he divided into any
two 2ici7'ts, the rectangle contained hy the lohole and one of. the
parts is equivalent to the square of that part^ together loith the
rectangle contained hy the tioo parts.

Let the straight line AB be divided into two parts in the
point C ; the rectangle AB.BC is equivalent to the square of
-BC, together with the rectangle AC.CB

Upon BC describe (I. 23) the square CE ; produce ED toF;

and through A draw (I. 18) AF parallel
to CD or BE. Then the rectangle AE
is equivalent (I. ax. 10) to the rectangles
CE, AD. But AE is the rectangle con-
tained by AB, BC, for it is contained
by x^B, BE, of which BE is equal to
BC ; and AD is contained by AC, CB,
for CD is equal to CB ; also DB is the square of BC. Thei-e-
fore the rectangle AB.BC is equivalent to the square of BC,


1 2 4 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

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 4 of 21)