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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 5 of 21)
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together with the rectangle AC.CB. If, therefore, etc.

8cho. Otherwise : Take a line D equal to CB. Then (II. 1)
the rectangle AB.D is equivalent to the

-^ ^ ^ rectangles BCD and AC.D ; that is (const.

and L def 15), the rectangle AB.BC is
^ equivalent to the square of BC togctlier with

the rectangle AC.CB.

Prop. IV.— Tiieor.— 7/" a straight line he divided into
any tico j^^-^^ts, the square of the v-hole line is equivalent to
the squares of the tioo i)arts\ together with twice their red

angle.

Let the straight line AB be divided into any two parts in C ;
the square of AB is equivalent to the squares of AC and CB,
together with twice the rectangle under AC and CB.

On AB describe (L 23) the square of AE, and join BD;
through C draw (I. 18) CGF parallel to AD or BE; and
through G draw UK parallel to AB or DE. Then, l)ecause



G




/



BOOK II.] EUCLID AND LEGENDEE. 51

CF is parallel to AD, aucl BD falls upon them, the exterior
angle CGB is equivalent (I. 16) to the inte-
rior and remote angle ADB ; but ADB is ^ ^ ?

•equal (I. l) to ABD, because BA and AD
are equal, being sides of a square; wherefore
(I. ax. 1) the angle CGB is equal to GBC;
iiad therefore the side BC is equal (I. 1, cor.)
to the side CG. But (const.) the figure CK » ^ B

is a parallelogram ; and since CBK is a right
angle, and BC equal to CG, CK (I. def 15) is a square, and it
is upon the side CB. For the like reason HF also is a square,
and it is upon the side HG, which is equal (I. 15, cor. 1) to AC;
therefore HF, CK are the squares of AC, CB. And because
(I. 15, cor. 8) the complements AG, GEare equivalent, and that
AG is the rectangle contained by AC, CB, for CG has been
proved to be equal to CB ; therefore GE is also equivalent to
the rectangle AC.CB; wherefore AG, GE are equivalent to
twice the rectangle AC.CB. The four figures, therefore, HF,
-CK, AG, GE are equivalent to the squares of AC, CB, and
twice the rectangle AC.CB. But HF, CK, AG, GE make up
the Avhole figure AE, which is the square of AB ; therefore the
square of AB is equivalent to the squares of AC and CB, and
twice the rectangle AC.CB. Wherefore, etc.

Otherwise: AB^ = AB.AC + AB.CB (H. 2). But (H. 3)
A.B. AC =AC=+ AC.CB, and AB.CB=zCB-+AC.BC. Hence
<I. ax. 2) AB.AC + AB.BC, or AB-=AC^+CB-+2AC.CB.

Cor. 1. It follows from this demonstration, that the parallel-
ograms about the diagonal of a square are likewise squares.

Cor. 2. Hence the square of a straight line • is equivalent to
four times the square of its half; for the straight line being
iDisected, the rectangle of the parts is equivalent to the square
of one of them.

Pkop. V. — Theor. — If a straight line be divided into two
' equal parts, and also into tioo unequal parts ; the rectangle con-
tained by the unequal parts., together with the square of the line
hettoeen the points of section, is equivalent to the square of half
the line.

Let the straio;ht line AB be bisected in C, and divided un-





H


/


K L


/



52 THE ELEMENTS OF [bOOK II.

equally in D ; the rectangle AD.DB, together with the square
of CD, is equivalent to the square of CB.

On CB describe (I. 23) the square CF, join BE, and through
D draw (I. 18) DHG parallel to CE or BF ; also through H

draw KLM parallel to CB or EF ; and
■^ c D B through A draw AK parallel to CL or

BM. Then (I. 15, cor. 5) AL and CM
^^ are equal, because AC is equal to CB ;

and (I. 15, cor. 8) the complements CH
E G F and HF are equivalent. Therefore (I.

ax. 2) AL and CH together are equal to
CM and HF together ; that is, AH is equivalent to the gnomon
CMC To each of those add LG, and (I. ax. 2) the gnomon
CMG, together with LG, is equivalent to AH together with
LG. But the gnomon CMG, and LG make up the figure
CEFB, which is the square of CB ; also AH is the rectangle
imder AD and DB, because DB is equal (IL 4, cor. 1) to DH ;
and LG is the square of CD. Thei-efore the rectangle AD.DB
and the square of CD are equivalent to the square of CB.
Wherefore, if a straight line, etc.

Otherwise : Since, as is easily seen from the proof in the
above, DF is equal to AL, take these separately from the entire
figure, and there remain AH and LG equivalent to the square
CF, as before. The proof may also be as follows:

AD.DB = CD.DB + AC.DB (H. 1) or AD.DB =CD.DB+
CB.DB, because CB=:AC. Hence (H. 3) AD.DB =CD.DB +
CD.DB+DB^=i:2CD.DB + DBl To each of these add CD=;
then AD.DB + CD^=2CD.DB+DB^+CD^ or (H. 4) AD.DB
+ CD'=CB\

Cor. 1. Hence the ditference of the squares of CB and CD is
equivalent to the rectangle under AD and DB. But since AC
is equal to CB, AD is equivalent to the sum of CB and CD, and
DB is the difterence of these lines. Hence the difference of the
squares of two straight lines is equivalent to the rectangle under
their sum and difference.

Cor. 2. Since the square of CB, or, which is the same, the
rectangle AC.CB, is greater than the rectangle AD.DB by the
equare of CD, it follows, that to divide a straight line into two



BOOK n.] EUCLID AND LEGENDKE. 53

parts, the rectangle of wliich may be the greatest possible, or,
as is termed, a tnaxlmiim, tlie line is to be bisected.

Cor. 3. Hence also tlie sum of the squares of tlie two parts
into which a straight line is divided, is the least possible, or is,
as it is termed, a ininimuni, Avlien the line is bisected. For (II.
4) the square of tlie line is equivalent to the squares of the
parts and twice their rectangle ; and therefore the greater the
rectangle is, the less are the squares of the j^arts; but, by the
foregoins; corollarv, the rectangle is a maximum when the line
is bisected.

C<yr. 4. Since (I. 24, cor. 5) the difference of the squares of
the sides of a triangle is equivalent to the difference of the
squares of the segments of the base, it follows, from the first
corollary above, that the rectangle under the sum and differ-
ence of the sides of a triangle is equivalent to the rectano-le
under the sum and difference of the segments, intercepted be-
tween the extremities of the base and the jjoint in which the
perpendicular cuts the base, or tlie base produced.

Cor. 5. Hence, also, if a straight line be drawn from the ver-
tex of an isosceles triangle to any point in the base, or its con-
tinuation, the difference of the squares of that line and either of
the equal sides is equivalent to the rectangle under the seoments
intercepted between the extremities of the base and the jDoint.

Cor. 6. Since (I. 24, cor. 4) the square of one of the legs of a
right-angled triangle is equivalent to the difference of the
squares of the hypothenuse and the other leg, it follows (IT. 5,
cor. 1) that the square of one leg of a right-angled triangle is
equivalent to the rectangle under the sum and difference of the
hypothenuse and the other.

^cho. And a parallelogram can be constructed equivalent
to a ffiven triangle, or anv given rectilinear fiorure havinor an
angle equal to a given angle, or applied to a given straight line
— that is, having that straight line for one of its sides, when
the jjarallelogram shall be equivalent to a given triangle or
given rectilinear figure, and have one of its angles equal to a
given angle, by applying a parallelogram equivalent to the
given triangle with an equal angle, to a given straight line, and
then constructing an equal triangle to the given triangle (I. 15
and 15, cor. 4).



A C


B D


K




H


/


L


/





54 THE ELEMENTS OF [bOOK II,

Prop. VI. — Theok. — If a straight line he bisected, and be
produced to any point, the rectangle contained by the whole
line thus produced, and the part of it produced, together with
the square of half the line bisected, is equivalent to the square
of the straight line which is made %ip of the half and the p>art
produced.

Let the straight line AB be bisected in C, and produced to
D ; the rectangle AD.DB, and the square of CB, are equiva-
lent to the square of CD.

Upon CD describe (I. 23) the square CF, and join DE ;
through B draw (I. 18) BHG parallel to CE or DF ; through

H draw KLM parallel to AD or EF, and
through A draw AK parallel to CL or
jj DM. Then because AC is equal to CB,
the rectangle AL is equal (I. 15, cor. 5) to
CH ; but (1.15, cor. 8) the complements CH,
J, ^j J, HF are equivalent ; therefore, also, AL is

equal to HF, To each of these add CM
and LG ; therefore AM and LG are equivalent to the whole
square CEFD, But AM is the rectangle under AD and DB,
because (11. 4, cor. 1) DB is equal to DM; also, LG is the
square of CB, and CEFD the square of CD. Therefore the
rectangle AD.DB and the square of CB are equivalent to the
square of CD. Wherefore, if a straight line, etc.

Otherwise : Produce CA to N, and make CN equal to CD.

To these add the equals CB and CA ;
^ ^ ^ ^ ° therefore NB is equal (L ax. 2) to AD.

But (IL 5) the rectangle NB.BD, oi-
AD.BD, together with the square of CB, is equivalent to the
square of CD ; which was to be proved.

Prop. VIL — Theor. — If a straight line be divided into a?2y
two parts, the squares of the whole line and of one of the p>arts
are equivalent to ticice the rectangle contained by the whole and
that part, together tcith the square of the other part.

Let the straight line AB be divided into any two parts in the
point C ; the squares of AB, BC are equivalent to twice the
rectangle AB.BC, together with the square of AC.



BOOK II.] EUCLID AND LEGENDRE. 65

Upon AB describe (I. 23) the square AE, and construct the
figure as in the preceding propositions. Then,
because (I. 15, cor. 8) the complements CII, ^ c B

FK are equivalent, add to each of thorn CK;
the whole AK is therefore equal to the whole ^
CE ; therefore AK, CE are together double
of AK. But AK, CE are the gnomon AKF,



K



together with the square CK ; therefore the i> F E

gnomon AKF and the square CK are double
of AK, or double of the rectangle AB.BC, because BC is equal
(II. 4, cor. 1) to BK. To each of these equals add HF, which
is equal (II. 4, cor. 1 ) to the square of AC ; therefore the gno-
mon AKF, and the squares CK, HF are equivalent to twice the
rectangle AB.BC and the square of AC. But the gnomon
AKF, and the squares CK, HF make up the whole figure AE,
together with CK; and these are the squares of AB and BC ;
therefore the squares of AB and BC are equivalent to twice the
rectangle AB.BC, together with the square of AC ; wherefore,
if a straight line, etc.

Otherwise: Because (II. 4) AB^ = AC=+BC-+2AC.BC, add
BC= to both; then AB=+BC==AC= + 2BC=+2AC.BC. But
(II. 3) BC=+AC.BC= AB.BC, and therefore 2BC=+2AC.BC
=2AB.BC; wherefore AB-+BC^=AC=+2AB.BC.

Cor. 1. Since AC is the diflierence of AB and BC, it follows
that the square of the difterence of two straight lines is equiva-
lent to the sum of their squares, wanting twice their rectangle.

Cor. 2. Since (II. 4) the square of the sum of two lines ex-
ceeds the sum of their squares by twice their rectangle, and
since, by the foregoing corollary, the square of their difference
is less than the sum of their squares by twice their rectangle, it
follows that the square of the sum of two lines is equivalent to
the square of their difference, together with four times their
rectangle.

Pkop. VIII. — TiiEOR. — If a straight line he divided into two
equal, and also into tvw imequal parts, the squares of the un-
equal parts are together double of the square of half the linCj
and of the square of the line hetioeen the points of section.

Let the straight line AB be divided equally in C, and un-



56



THE ELICMENTS OF



[book n.



E



equally in D ; the square? of AD, DB are together double of
the squares of AC, CD.

From C draAV (I. 7) CE j)ei-|ie!Klicular to AB, and make it
equal to AC or CB, and join EA, EB ; through D draw (I. 18)
DF jjarallel to CE, and tlirougli V draw FG parallel to AB ;
and join AF. Then because (const.) the triangles ACE, BCE
are right-angled and isosceles, tlie angles CAE, AEC, CEB,

O O 7 O 7 3 7

EBC are (I. 20, cor. 4) each half a riglit angle. So also (I. 16,
part 2) are EEC, BED, because FG is parallel to AB, and ED
to EC ; and for the same reason EGF, ADF are right angles.
The angle AEB is also a right angle, its parts being each half
a right angle. Hence (I. 1) EG is equal to GF or CD, and ED

to DB. Again (I. 24, cor. 1) : the square
of AE is equivalent to the squares of AC,
CE, or to tAvice the square of AC, since
AC and CE are equal. In like manner,
the square of EF is equivalent to twice
the square of GF or CD. Now (I. 24,
^ cor. 1), the squares of AD and DF, or
of AD and DB, are equivalent to the square of AF ; and the
squares of AE, EF, that is, twice the square of AC and twice
the square of CD, are also equivalent to the square of AF;
therefore (I. ax. 1) the squares of AD, DB are equivalent to
twice the square of AC and twice the square of CD. If, there-
fore, a straight line, etc.

Otherwise: DB^+2BC.CD=BC=+CD^ (II. 7), or DB=+
2AC.CD=:AC= + CD^-; also (11. 4) AD==:AC^+CD=+2AC.
CD. Add these equals togethei-, and from the sums take
2AC.CD; then AD-+DB^=2AC^+2CD^




Prop. IX. — ^Theor. — If a straight line he bisected, and pro-
duced to any point, the squares of the whole line thus produced,
and of the part of it produced, are together double of the square
of half the lifie bisected, and of the square of the line made up
of the half and the part produced.

Let the straight line AB be bisected in C, and produced to
D ; the squares of AD, DB arc double of the squares of AC,
CD.

From C draw (I. 7) CE perpendicular to AB ; and make it



BOOK II.]



EUCLID AND LEGENDRE.



57



A




equal to AC or CB ; join AE, EB, and tln-ough E and D draw
(I. 1 8) EF parallel to AB, and
DF parallel to CE. Then be-
cause the straiu'lit line EF
meets the parallels EC, FD,
the ano-les CEF, EFD are
equivalent (I. 9) to two right
anixles : and therefore the an-
gles BEF, EFD are less than

two right angles; therefore (I. 19) EB, FD will meet, if pro-
duced toward B, D ; let them meet in G, and join AG. Then
it would be proved, as in the last proposition, that the angles
CAE, xVEC, CEB, EBC are each half a right angle, and AEB
a right angle. BDG is also a right angle, being equal (I. 16)
to ECB, since (const.) EC, EG are parallel ; DGB, DBG are
each half a right angle, being equal (I. 10 and I. 11) to CEB,
CBE, each to each ; and FEG is half a right angle, being (I.
16) equal to CBE. It would also be proved, as in the last
proposition, that the squai-e of AE is twice the square of AC,
and the square of EG twice the square of EF or CD. Now
(I. 24, cor. l),the squares of AD, DG, or of AD, DB, are equiv-
alent to tlie sipiare of AG; and the squares of AE, EG, or twice
the square of AC and twice the square of CD, are also equiva-
lent to the square of AG. Therefore (I. ax. 1) the squares of
AD, DB are equivalent to twice the square of AC and twice
the square of CD. If, therefore, a straight line, etc.

Otherwise: Produce CA, making CH equal to CD. To
these add CB, CA ; therefore IIB, AD are
equal. Then (II. 8) IIB - +BD^ or AD=+
BD^ = 2CD=-f2AC-.

&eho. The nine foregoing propositions may all be proved
very easily by means of algebra, in connection with the princi-
ples of mensuration, already established in the corollaries to
the 23d proposition of the first book. Thus, to pi'ove the fourth
proposition, let AC=:a, CB = 5, and, consequently, AB=a+5.
Now, the area of the square described on AB will be found (I.
23, cor. 4) by multiplying a-\-h by itself. This product is
found, by performing tlie actual operation, to be «'+2aJ+^';
an expression, the first and third parts of which are, by the



n A



B D



58 THE ELEMENTS OF [bOOK H.

same corollary, the areas of the squares of AC and CB, and the
second is twice the rectangle of those lines.

In like manner, to prove the eighth, adopting the same nota-
tion, Ave have the line which is made up of the whole and CB
= a + 2$y and, multiplying this by itself, we get for the area ot
the square of that line, cr+iab+ib", or a'-\-4{a-]-b)b, the
first part of which is the area of the square of AC, and the
second four times the area of the rectangle under AB and CB.

It will be a useful exercise for the student to prove the other
propositions in a similar manner. He will also find it easy to
investigate various other relations of lines and their parts by
means of algebra.

All the properties delivered in these propositions hold also
respecting numbers, if products be substituted for rectangles.
Thus, 7 being equal to the sum of 5 and 2, the square, or sec-
ond power of 7, is equal to the squares of 5 and 2 and twice
their product ; that is, 49 = 25 + 4+20.

Pkop. X. — Peob. — To divide a given straight line into txoo
parts, so that the rectangle contained by the \chole and one of
the parts may be equivalent to the square of the other part.

Let AB be the given straight line ; it is required to divide it
into two parts, so that the rectangle under the whole and one
of the parts may be equivalent to the square of the other.

Upon AB describe (I. 23) the square AD ; bisect (I. 6) AC
in E, and join E with B, the remote extremity of AB ; produce
CA to F, making EF equal to EB, and cut off AH equal to AF ;
AB is divided in H, so that the rectangle AB.BH is equivalent
to the square of AH.

Complete the parallelogram AG, and produce GH to K.
Then, since BAC is a right angle, FAH is also (I. 9) a right
angle; and (I. def 15) AG is a square, AF, AH being equal
by construction. Because the straight line AC is bisected in
E, and produced to F, the rectangle CF.FA and the square of
AE are together equivalent (II. 6) to the square of EF or of EB,
since (const.) EB, EF are equal. But the squares of BA, AE
are equivalent (I. 24, cor. 1) to the square of EB, because the
angle EAB is a right angle; therefore the rectangle CF.FA
and the square of AE are equivalent (I. ax. 1) to the squares of



BOOK n.]



EUCLID AND LEGENDRE.



59



BA, AE. Take away the square of AE, -u-hich is common to
both; therefore the remainino- rectano-le CF.FA



F



G



E





H


-"





B



D



is equivalent (I. ax. 3) to the square of AB.

But the figure FK is the rectangle contained

by CF, FA, for AF is equal to FG ; and AD

is the square of AB ; therefore FK is equal

to AD. Take away the common part AK,

and (I. ax. 3) the remainders FH and HD are

equivalent. But HD is the rectangle AB.BH,

for AB is equal to BD ; and FH is the square

of AH. Therefore the rectangle AB.BH is equivalent to the

square of AH ; Avherefore the straight line AB is divided in H,

so that the rectangle AB.BH is equal to the square of AH ;,

which was to be done.

Sc/w. 1. In the practical construction in this proposition, and
in the 2d cor. to 9 of the fifth book, which is virtually the same, it
is sufficient to draw AE perpendicular to AB, making it equal
to the half of AB, and producing it through A ; then, to make
EF equal to the distance from E to B, and AH equal to AF.
It is plain that BD might be bisected instead of AC, and that
in this way another point of section would be obtained.

While the enunciation in the text serves for ordinary pur-
poses, it is too limited in a geometrical sense, as it comprehends
only one case, excluding another. The following include&
both :

In a given straight line^ or its continuatio7i, to find ai^ointy
such that the rectangle contained hy the given line^ and the seg-
ment between one of its extremities and the required point, may
be equal to the square of the segment between its other extremity
and the same point.

The point in the continuation of BA will be found by cutting
ofiT a line on EC and its continuation, equal to EB, and describ-
ing on the line composed of that line and AE a square lying on
the opposite side of AC from AD ; as the angular j^oint of that
square in the continuation of BA is the point required. The
proof is the same as that given above, except that a rectangle
corresponding to AK is to be added instead of being sub-
tracted.

Scho. 2. The line CF is equivalent to BA and AH ; and since



60 THE ELEMENTS OF [bOOK H,

it has been shown that the rectangle CF.FA is eqiiivalent to
the square of BA or CA, it follows that if any straight line AB
(see the next diagram) be divided according to this proposition
in C, AC being the greater part, and if AD be made equal to
AB, DC is similarly divided in A. So also if DE be made



F



E D A C B



equal to DC, and EF to EA, EA is divided similarly in D, and
FD in E ; and the like additions may be continued as far as we
please.

Conversely, if any straight line FD be divided according to
this proposition in E, and if EA be made equal to EF, DC to
DE, etc., EA is similarly divided in D, DC in A, etc. It fol-
lows also, that the greater segment of a line so divided will be
itself similarly divided, if a part be cut oif from it equal to the
less ; and that by adding to the whole line its greater segment,
another line will be obtained, which is similarly divided.

Prop. XI. — Theok. — In an oUuse-angkd triangle^ the
square of the greatest side exceeds the squares of the other tioo^
hy tioice the rectangle contained by either of the last-mentioned
sides, and its continuation to meet a perpendicular drawn to it
from the opposite angle.

Let ABC be a triangle, having the angle ACB obtuse ; and
let AD be perpendicular to BC produced ; the square of AB is
equivalent to the squares of AC and CB, and twice the rectan-
gle BC.CD.

Because the straight line BD is divided into two parts in the

point C, the square of BD is equivalent (II. 4) to the squares

of BC and CD, and tAvice the rectangle

BC.CD. To each of these equivalents add

the square of DA ; and the squares of DB,

DA are equivalent to the squares of BC,

CD, DA, and twice the rectangle BC.CD.

But, because the angle D is a right angle,

the square of BA is equivalent (I. 24, cor. 1) to the squares of

BD, DA, and tlie square of CA is equivalent to the squares of




BOOK II.] EUCLID AND LEGENDRE. 61

CD, DA ; therefore the square of BA is equivalent to the
squares of BC, CA, and twice the rectangle BC.CD. Therefore,
in an obtuse-angled triangle, etc.

Pkop. XII. — TiiEOR. — I/i any triangle^ the square of a side
subtending an acute a?igle is less than the squares of the other
sides, by twice the rectangle contained by either of those sides,
and the straight line hitercej^ted between the acute angle and
the perpendicular drawn to that side from tJie opposite angle.

Let ABC (see this figure and that of the foregoing proposi-
tion) be any triangle, having the angle B acute ; and let AD
be perpendicular to BC, one of the sides containing that angle;
the square of AC is less than the squares of AB, BC, by twice
the rectangle CB.BD.

The squares of CB, BD are equivalent (II. 1) to twice the
rectangle contained by CB, BD, and the square of DC. To
each of these equals add the square of
AD ; therefore the squares of CB, BD,
DA are equivalent to twice the rect-
angle CB.BD, and the squares of AD,
DC. But, because AD is perpendicu-
lar to BC, the square of AB is equiva-
lent (I. 24, cor. 1) to the squares of

BD, DA, and the square of x\.C to the squares of .VD, DC ;
therefore the squares of CB, BA are equivalent to the square of
AC, and twice the rectangle CB.BD ; tliat is, tlie square of AC
alone is less than the squares of CB, BA, by twice the rectangle
CB.BD.

If the side AC be perpendicular to BC, then BC is the
straight line between the perpendicular and the acute angle at
B ; and it is manifest that the squares of AB, BC are equiva-
lent (I. 24, cor. 1) to the square of AC and twice the square of
BC. Therefore, in any triangle, etc.

Scho. By means of this or the foregoing proposition, the area
of a triangle may be computed, if the sides be given in num-
bers. Thus, let AB = 17, BC = 28, and AC = 25. From AB=+
BC take AC; that is, from l7'4-2SUake 25=; the remainder
448 is twice CB.BD. Dividing this by 56, twice BC, the quo-
tient 8 is BD, Hence, from either of the triangles ABD,




62



THE ELEMENTS OF [bOOK II.



ACD, we find the perpendicular AD to be 15 ; and thence the
area is found, by taking half tlie product of BC and AD, to be

210.

The segments of the base are more easily found by means -of
tJie 4th corollary to the fifth proposition of this book, in connec-
tion with the pi-inciple, that if half the difference of two mag-
nitudes le added to half their sum ^ the resvlt is the greater; and
if half the difference he taken from half the sum, the remainder


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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 5 of 21)