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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This proposition is the converse of the third proposition, and
is susceptible of three cases, viz. : first, when the equal sides are
between the equal angles ; secondly and thirdly, when the equal
sides are opposite to the equal angles similarly situated.

Q J, Let ABC and DEF be two triangles

M-hich have the angles ACB and EFD
equal; the angles BAG and DEF
equal, and the sides CA and FE equal ;
then the sides CB and FD are equal,
also the sides AB and DE ; the angles
CBA and FDE are equal, and the tri-
angles are equal to one another.

If the triangles be placed so as to
have their sides CB and FD in the same straight line, but the
triangles be on opposite sides of that line, and the vertex C on

the vertex F ; then because the angles
AFD and EFD are equal (hyp.), the angle
AFE is bisected by FD ; and because AF
and FE are equal (hyp.), the triangle AFE
is isosceles (I. def 13), and the line AE
(T. 6) is also bisected by FD; hence (I. 1,
cor. 1) the angles FAE and FEA are
equal ; therefore (I. 3) the triangles FAH
and FEII are equal. But the angles FAD

BOOK I.] EUCLID AND LEGENDKE. 27

and FED are eq^^al (byp.) ; tlien taking from each the equal
angles FAII and FEH, there will remain (I. ax, 3) the angle
HAD equal to the angle HED; hence (I. 1, cor. 2) the triangle
AED is isosceles, and (I. def. 13 and I. cor. 2) the side AD is
equal to the side ED, and AE being bisected by FD (I. 6), we
have the triangles AHD and EHD (I. 3) equal ; therefore the
angles HDA and HDE are equal; hence the triangles AFD
and EFD have the sides AF and EF (hyp.) equal, the angles
FAD and FED equal (hyp.), and the sides AD and DE equal
(I. def 13 and I, cor. 2) ; therefore (I. 3) the angles FDA and
FDE are equal ; the side FD is common, and the two triangles
are equal.

In a similar manner, the second and third cases can be dem-
onstrated. Wherefore, if two angles of one triangle be, etc.

Cor. Hence, from this proposition, it can also be shown that
the second corollary to the first proposition is true. Let the
triangle ABC have the angle CAB equal c

to the angle CBA, then will AC be equal
toCB.

From the vertex C (I. S) draw CD per-
pendicular to AB, then the angles CDA
and CDB are both right angles (I. def 10), ^
the angles CAD and CBD are equal (hyp.), and the side CD coni-
mon (const.); therefore (I. 14) the sides AC and CB are equal.

8cho. It will be seen from propositions third, fourth, and
fourteenth of this book, that two triangles are in every respect
equal when the three sides of the one are respectively equal to
the three sides of the other, when two triangles have two angles
and a side in each equal, each to each, and when one angle and
two sides of one are equal to one angle and two sides of the
other, each to each, and the equal angles in each triangle simi-
larly situated with respect to those sides, but with this limita-
tion, that when two equal sides respectively of the triangles are
the least sides, that the angles opposite the greater sides respec-
tively of the triangles must be of the same kind, either both
acute, both right-angled, or both obtuse. And from these prop-
ositions it is shown that of the sides and angles of a triangle,
three must be given to determine the triangle, and these three
can not all be angles. Were only three angles given, the sides,

28 THE ELEMENTS OF [bOOK I.

as will appear from the twentieth prosposltion of this book,
might be of any magnitude whatever. The pupil may occupy
himself in proving these propositions by superposition or
some other way, as by pursuing a course of demonstration
different from what is given in the text, he will more readily
familiarize himself with the process of geometrical reasoning.

Prop. XV. â€” Theor. â€” Parallel straight liries are equally/ dis-
tant from each other, however so far they be produced on the
same plane.

Let the straischt line AB
be bisected (I. 6) at C, and
let the perpendicular CD be
drawn (I. 7) ; join AD and
A c ' ^ BD, and if the triangle ADC

be applied to the triangle BDC so that they will fall on difler-
â‚¬nt sides of BD and have D and BD common, their sides DE
and CB will be equally distant from each other, and so they
will never meet, however so far they be produced on the same
plane, and consequently (I. def 11) are parallel straight lines.

Because DC is perpendicular to AB (const.), the angles ACD
and BCD are both right angles (I, def 10), and are equal (I. ax.
11) ; hence the triangles ADC and BDC have the side DC
common, the sides AC and CB equal (const.), and the angles
ACD and BCD equal (I. def 10 and ax. 11); therefore (I. 3)
the triangles are equal, having the sides AD and BD equal ;
the angle ADC equal to the angle BDC, and the angle DAC
equal to the angle DBC. Now, when ADC is applied to BDC
so that they will fall on difterent sides of BD, and have D and
BD common (hyp.), since AD is equal to BD, the point A will
fall on B ; hence the angle EBD will be the same as ADC, and
equal to BDC, the angle BED the same as ACD, and equal to
DCB, the angle EDB the same as DAC, and equal to DBC, the
side DE the same as AC, and equal to AC ; the side EB equal
to DC; therefore the triangles BDC and BDE are equal (I.
ax. 8), having their sides and angles equal, each to each.

Since DE and CB are straight lines, they have no variation
in the direction of their lengths from D to E or from C to B
(I. def 7) ; and because DC is equal to EB, DE and CB are

BOOK I.] EUCLID AND LEGENDEE. 29

equally distant from each other at their extremities, and having
no variation in the direction of their lengths from D to E and
from C to B (I. def. 7), they are also equally distant from each
other at every part between D and E and C and B, each to
each ; therefore DE and CB (I. post. 2) on being produced to
any length are still the same straight lines, and will have no
variation in the direction of their lengths (I. def. 7), conse-
quently they will always be the distant DC or EB from each
other, at every part, each to each ; and being always the same
distant DC or EB from each other, will never meet, and are
parallel straight lines (I. def 11). Wherefore, parallel straight
lines, etc.

Cor. 1. In like manner, it can be shown that DC and EB
are parallel lines; hence DCBE is a parallelogram (I. def 14) ;
and since the angles DCB and DEB are equal, and the angles
EDB and BDC equal to the angles EBD and DBC (I. ax. 2),
and the sides DE and CB equal, also the sides CD and BE
equal, a parallelogram has its opposite sides and opposite angles
equal.

Cor. 2. Hence parallel lines, DE and CB, intercepted by a
straight line, DB, make the alternate angles EDB and CBD
equal; and, co?itÂ»er5eZy, when the alternate angles EDB and CBD
are equal, the lines DE and CB are parallel.

Cor. 3. In the parallelogram, the opposite sides being equal,
the straight lines which join the extremities of two equal and
parallel straight lines toward the same parts â€” that is, the near-
est extremities together â€” are themselves equal and parallel ;
hence a quadrilateral which has two sides equal and parallel is
(I. def. 14) a parallelogram.

Cor. 4. Because the triangles DCB and DEB are equal, a
diagonal, DB, bisects the parallelogram ; and if two parallelo-
grams have an angle of the one equal to an angle of the other,
and the sides containing those equal angles respectively equal,
the parallelograms are equal, as the parallelograms can be bi-
sected by diagonals subtended by the equal angles, and the tri-
angles thus formed are equal (I. 3) ; hence (I. ax. 6) the par-
allelograms are equal; hence, also, if a pai'allelograra and a tri-
angle be upon the same or equal bases, and between the same
parallels, the parallelogram is double the triangle.

30 THE ELEMENTS OF [bOOK I.

Cor. 5. Hence, also, parallelograms upon the same or equal
bases, and between the same parallels, are equal ; and triangles
upon the same or equal bases and between the same parallels,
are equal.

Cor. 6. Hence, from the preceding corollary, it is plain that,
triangles or parallelograms between the same parallels, but
upon unequal bases, are unequal.

Cor. T. And a straight line drawn from the vertex of a tri-
angle to the point of bisection of the base, bisects the triangle ;
and if two triangles have two sides of the one respectively equal
to two sides of the other, and the contained angles supple-
mental (I. def 20), the triangles are equivalent; the converse is
also true.

Cor. 8. If through any point in either diagonal of a parallelo-
gram straight lines be drawn parallel to the sides of the four
parallelograms thus formed, those through which the diagonal
does not pass, and which are called the com2)lements of the
other two, are equivalent.

Peop. XVI. â€” Theoe. â€” If a straight linefallupon tioo parallel
straight Ihies, (l) it makes the alternate angles equal to one an-
other ; (2) the exterior angle equal to the interior and remote
upon the same side, and (3) the tico interior angles upon the
same side together, equivalent to two right angles.

Let the straight lines AB, CD be parallel, and let EF fall upon
them; then (l) the alternate angles AGH, GHD are equal to
one another ; (2) the exterior angle EGB is equal to the interior
and remote upon the same side, GHD; and (3) the two in-
terior angles BGH, GHD ujDon the same side are together
equivalent to two right angles.

Since AB, CD are parallel (hyp.), theper-

B pendiculars (I. 7) MG, LH make the angles

MGH, GIIL equal to one another (I. 15,

cor. 2), and AG3I, LHD are two riglit

.jj angles (I. def 10); if Ave add AGM to

MGH they will be equal to AGH (I. ax.

^' 10), and DHL added to LIIG are likewise

equal to GIID ; hence (I. ax. 2) AGH and GHD are equal to

one another.

BOOK I.] EUCLID AND LEGENDRE. 31

Second. AGII is equal to EGB (I. 11), therefore (I. ax, 1)
EGB is equal to GHD.

Third. Add to EGB and GHD, each, the angle BGII ; there-
fore (I. ax. 2 and ax. 10) EGB and BGH are equivalent to the
angles GHD and BGH, but EGB and BGH are equivalent to
two right angles (I. 9) ; therefore, also, BGH and GHD are to-
gether equivalent to two right angles. Wherefore, if a straight
line, etc.

Cor. 1. Hence, conversely, tAvo straight lines are parallel to
one another, if another straight line falling on them (1) makes
the alternate angles equal ; (2) the exterior angle equal to the
interior and remote upon the same side of that line ; and (3)
the two interior angles upon the same side together equivalent
to two right angles.

Let EF fall on AB and CD, the perpendiculars (I. 1) GM and
HL make two riglit angles (I. def. 10), AGM and DHL. But
AGH and DHG are equal (hyp.) ; hence (L ax. 3) MGH is
â– equal to LHG. But MGL and LHM are both right angles
(const, and L def. 10) ; hence (I. ax. 3) HGL is equal to GHM;
â€˘therefore in the triangles HMG and GLH we have two angles
in one equal to two angles in the other, each to each, and the
side GH common, and (I. 14) the triangles are equal; hence
GM and HL are equal, and (L 15) AB and CD are equally dis-
tant from each other, and will never meet on being produced
{L post. 2), and are parallel (L def' 11).

Because EGB is equal to DHG (hyp.), and EGB equal to
AGH (I. 11), the angle AGH is equal to DHG (L ax. 1), but
they are alternate angles; therefore (L 16, cor. 1, part l) AB is
parallel to CD. Again : because BGH and GHD are together
equivalent to two right angles (hyp.), and AGH and BGH are
also equivalent to two right angles (I. 9), the angles AGH and
BGH are together equal to BGH and GHD (I. 7 and ax. 1) ;
then (L ax. 3) AGH is equal to GHD, but they are alternate
angles, therefore (I. 16, cor. 1, part l) AB and ED are parallel.
Wherefore, two straight lines are parallel, etc.

Cor. 2. When one angle of a parallelogram is a right angle,
all the other angles are right angles; for since (I. 10) BGM and
GMH are together equivalent to two right angles, if one of
them be a right angle, the other must also be a right angle, and

32 THE ELEMENTS OF [boOK I.

(I. 15, cor. 1) the opposite angles are equal. A rectangle, then
(L def. 15), has all its angles right angles.

Cor. 3. If two straight lines make an angle, two others par-
allel to them contain an equal or supplemental angle ; thus LGM,
and the vertical angle produced by AB and the continuance of
MG through G, are each equal to LHM, while the angle AGM,
and its vertical angle contained by GB and the continuance of
MG, are each equal to the supplement of LHM ; hence we can
divide a given straight line AB into any proposed number of
equal parts.

Prop. XVH. â€” Theor. â€” Two straight lines which are not
in the same straight line, and which nre parallel to a third
straight line, are parallel to one another.

Let the straight lines AB, CD be each of them parallel to
the straight line EF ; AB is also parallel to CD.

Let the straight line LH cut AB, CD, EF ; and because LH
L cuts the parallel straight lines AB, EF, the
angle LGB is equal (L 16, part 2) to the
. J} angle LHF. Again : because the straight
line LH cuts the parallel straight lines CD,
" ^ EF, the angle LKD is equal (L 1 6, part 2) to

^ â–  " â€” J. the angle LHF ; and it has been shown

that the angle LGB is equal to LHF;

wherefore, also, LGB is equal (L ax. 1) to

LKD, the interior and remote angle on the same side of LH ;

therefore AB is parallel (L 10, part 1) to CD. Wherefore, two

straight lines, etc.

Prop. XVIIL â€” Prob. â€” To draw a straight line parallel to a
given straight line through a given poi^it without it.

Let AB be the given straight line, and C the given point ; it
is required to draw a straight line through C, parallel to AB.

In AB take any point D, and join CD ; at the point C, in the

c straight line CD, make (I. 13) the angle

E â€” -^ F p^j, ^^^^^^ ^^ ^j^j, ^ ^^^ produce the

A ^ B Straight line EC to any point F.

Because (const.) the straight line CD,
which meets the two straight lines AB, EF, makes the alternate

BOOK I.] EUCLID AND LEGENDEE. 33

angles ECD, CDB equal to one another, EF is parallel (I.
15, eor, 2) to AB. Therefore the straight line EOF is drawn
through the given point C parallel to the given straight linev
AB ; which was to be done. â€˘

Prop. XIX, â€” Theok. â€” If a straight line meet two otho'^
straight lines xohich are in the same pkoie, so as to make the
two interior angles on the same side of it, taken together, less
than two right angles, these straight lines shall at length meet
upon that side, if they he continually i^i'oduced. Axiom
twelfth. Elements of Euclid.

Let EF be the straight line meeting AB, ^
CD on the same plane, so tliat BLO, DOL
are less than two right angles, the lines
AB, CD will meet if continually produced.

At the point O, draw GH (I. 1 8) parallel g
to AB, then BLO, HOL are equivalent to
two right angles (L IG), but DOL is less c

than HOL (I. ax. 9) ; therefore BLO, DOL are less than BLO,
HOL, and less than two right angles. But GH, which fornij^
with EF the angle HOL, is parallel (const.) with AB, which!
forms with EF the angle BLO, and (L def 11) GH and AB-
never meet each other, because (L 15) they are equally distant
from each other, and (L 16, cor. l) the interior angles HOL^
BLO together equivalent to two right angles ; therefore.^
then, CD, which forms Avith EF the angle DOL less than HOL,
can not be parallel with AB (L ax. 9) ; and not being parallel
wath AB, CD and AB can not preserve the condition of par-
allel lines (L def 11), and will meet. And the line CD making
with EF the angle DOL less than HOL, on the side of EF,
where the interior angles BLO, DOL are less than two ricrht
angles, the line OD must therefore be between GH and AB,
and is consequently nearer to AB than GH is to AB (L ax. 9) ;
but CD making with EF the angle EOC greater than the angle
GOE, which GH makes with EF, therefore CO must be with-
out AB and GH, and consequently is farther from AB than GH
is from AB ; hence the straight line CD, Avhich is made of CO
and OD, has parts unequally distant from AB ; therefore since
CD approaches AB on the side of EF where OD is, CD must
3

S^ THE ELEMENTS OF [bOOK I.

meet AB on that side wliea continuallj^ produced, but OD
makes the angle DOL less than HOL, therefore CD will meet
AB on the side of EF where the angles BLO, DOL are less
than BLO, HOL. Wherefore, if a straight line, etc.

Cor. 1. Hence a straight line which intercepts one of two or
more parallel straight lines will intercept the others if continu-
ally produced ; hence, also, two straight lines which intercept
each other are not both parallel to the same straight line.

Prop. XX. â€” Theor. â€” If a side of any triangle be produced,
(1) the exterior angle is equivalent to the tico interior and re-
mote angles ; and (2) the three interior angles of every triangle
are together equivalent to two right angles.

Let ABC be a triangle, and let one of its sides BC be pro-
educed to D; (1) the exterior angle ACD is equivalent to the
two interior and remote angles CAB, ABC; and (2) the three
interior angles, ABC, BCA, CAB, are together equivalent to
two right angles.

Through tlae point C draw (L 18) CE parallel to the straight

line AB. Then, because AB is parallel to EC, and AC falls

'>iipon them, the alternate angles BAC, ACE are (L 16, part 1)

equal Again: because AB is ijarallel to
EC, and BD falls upon them, the exterior
angle ECD is equivalent (L I G, part 2)
to the interior and remote ano-le ^VBC :

'B^ ^ â€” D but the angle ACE has been sliown to

F be equal to the angle BAC ; there-
fore the whole exterior angle ACD is equivalent (I. ax. 2) to
the two interior and remote angles CAB, ABC. To these
equals add the angle ACB, and the angles ACD, ACB are
equivalent (L ax. 2) to the three angles CBA, BAC, ACB ; but
the angles ACD, ACB are equivalent (I. 9) to two right angles;
therefore, also, the angles CBA, BAC, ACB are equivalent to
two right angles. Wherefore, if a side, etc.

Another proof of this important proposition can be given by
producing the side AC through C to F. !N"ow (I. 11), the angle
DCF is equal to the angle ACB ; EC being parallel (const.) to
BA, the exterior angle ECD is equal to the intei'ior and remote
angle ABC (L 10), and for the same reason the alternate angles

r

BOOK I.J EUCLID AND LEGENDRE. 35

EGA and CAB are equal; hence we have the three angles of
the triangle, ACB, CBA, and CAB, equal to the three angles
DCF, DCE, and EGA, each to each ; but (I. 9) tlie angles DCF,
DCE, and ECA are equivalent to two right angles ; therefore
(I. ax. 1) the three angles ACB, CBA, and CAB of the triangle
are likewise equivalent to two right angles. And when a par-
allelogram (I. def 14) is formed by drawing (I. 18) pai-allels to
BA and AC respectively, it can be shown by the sixteenth
proposition tliat two adjacent angles of the parallelogram are
equivalent to two right angles, and the four angles together
equivalent to four right angles; since (I. 15, cor. 4) a diagonal
bisects the parallelogram and forms two equal triangles, the
angles are also equally divided, hence each triangle has its
three angles equivalent to two right angles.

Cor. 1. All tlie interior angles of any rectilineal figure, to-
gether Avith four right angles, are equivalent to twice as many
right angles as the figure has sides.

For any rectilineal figure can be divided into as many tri-
angles as the figure has sides, by draAving straight lines from a
point within the figure to each of its angles; and by the prop-
osition, all the angles of these triangles are equivalent to twice
as many right angles as there are triangles â€” that is, as there are
rsides of the figure ; and the same angles are equivalent to the
angles of the figure, together with the angles at the point Avhich is
the common vertex of the triangles â€” that is (I. 9, cor.), together
with four right angles. Tlierefore all the angles of the figure,
together Avitli four i-ight angles, are equivalent (I. ax. 1) to
twice as many riglit angles as the figure has sides.

Scho. 1. Another proof of this corollary may be obtained by
â– dividing the figure into triangles by lines drawn from any
.ano-les to all the remote angles. Then each of the tAVO ex-
treme triangles has tAVO sides of the polygon for tAvo of its sides,
while each of the other triangles has only one side of the figure
for one of its sides ; and hence the number of triangles is less
by tAVO than the number of the sides of the figure. Biit the in-
terior angles of the figure are evidently equivalent to all the
interior angles of all the triangles â€” that is, to tAvice as many
right angles as there are triangles, or tAvice as many right
angles as the figure has sides, less the angles of two triangles â€”

36 THE ELEMENTS OP [bOOK I.

tlfet is, four riglit angles. Hence, iu any equiangular figure,
the number of the sides being known, the magnitude of each
angle compared with a right angle can be determined. Thus,
in a regular pentagon, the amount of all the angles being twice
five *right angles less fourâ€” that is, six right angles, each angle
will be one fifth j^art of six right angles, or one right angle and
one fifth. In a similar manner it would appear that in the
regular hexagon, each angle is a sixth part of eight right angles,
or a right angle and a third ; that in the regular heptagon, each
is a right angle and three sevenths ; in the regular octagon, a
rio-ht ano;le and a half, etc.

Cor. 2. All the exterior angles of any rectilineal figure are
together equivalent to four right angles.

Because each interior angle ABC, and the adjacent exterior
ABD, are together equivalent (I. 9) to
two right angles, therefore all the in-
terior, together with all the exterior
angles of the figure, are equivalent to
twice as many right angles as there are
B sides of the figure â€” that is, hj the fore-

going corollary, they are equivalent to all the interior angles
of the figure, together with four right angles; therefore all the
exterior angles are equivalent (I. ax. .3) to four right angles.

Sc/to. 2. It is to be observed, that if angles be taken in the
ordinary meaning, as understood by Euclid, this corollary and
the foregoing are not applicable when the figures have re-entrant

angles â€” that is, such as open outward.
The second corollary will hold, how-
ever, if the difference between each
re-entrant angle and tAvo riiiht ansrles
be taken from the sum of the other exterior anccles: and the
former will be applicable, if, instead of the angle which opens
externally, the difterence between it and four right angles be
used. Both corollaries, indeed, Avill hold without change, if the
re-entrant angle be regarded as internal and greater than two
right angles; and if, to find the exterior angles, the interior be
taken, in the algeljraic sense, from two right angles, as in this
case, the re-entering angles will give negative or subtractive
results.

BOOK I.] EUCLID AND LEGENDEE. 87

Cor. 3. If a triangle has a right angle, the remaining angles
are together eqiiivaleut to a right angle ; and if one angle of a
triangle be equivalent to the other two, it is a right angle.

Cor. 4. The angles at the base of a right-angled isosctles tri-
angle are each half a right angle.

Cor. 5. If two angles of one triangle be equal to two angles
of another, their remaining angles are equal.

Cor. 6. Each angle of an equilateral triangle is one third of
two right angles, or two thirds of one right angle.

Cor. 1. Hence, a right angle may be trisected by describing
an equilateral triangle on one of the lines containing the right
angle.

ScJio. 3. By this principle also, in connection with the fifth
proposition, we may trisect any angle, Avhich is obtained by the
successive bisection of a right angle, such as the half, the
fourth, the eighth, of a right angle, and so on.

Cor. 8. Any two angles of a triangle are less than two right
angles.

Cor. 9. Hence every triangle must have at least two acute
angles.

Peop. XXI. â€” Tiieor. â€” Iftxco sides of a triangle be unequal,
(1) the greater side has the greater angle opposite to it ; and (2)
conversely, if tioo angles of a triangle be unequal, the greater
angle has the greater side opjiosite to it.

D Let ABC be a triangle of which the side

AB is greater than the side AC ; the angle
ACB opposite AB is greater than the angle
ABC opposite AC.

Because AB is greater than AC, pro-
duce AC (I. post. 2), and with A as a center, and a radius AB,
describe a circle (I. post. 3) intercepting AC produced in D,
and join BD ; the triangle ADB is isosceles (I. defs. 13 and 16) ;