Charles Hutton.

A course of mathematics for the use of academies, as well as private tuition .. online

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of the two inward angles on the same side, less than two
right angles, those two lines will not be parallel, but will
meet each other when produced.

THEOREK XV.

Those lines which are parallel to the same line, are
parallel to each other.

Let the lines ab, cd, be each of them G y >
parallel to the line bf ; then shall the lines -A. • o

AB, CD, be parallel to each other. C xj! 3D

For, let the line gi be perpendicular v \ p

to EP. Then will this line be also per- ^ i ^

pendicular to both the lines ab, cd (corol. th. 12), and con-
sequently the two lines ab, cd, are parallels (corol. th. 13).

a« B- n«

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THBOBSM XVI.

Whek erne side of a triangle is produced^ the ontmunl
angle is equal to both the inward opposite angles takeo
together.

Let the side, ab, of the triangle
ABC, be produced to d ; then will the
outward angle cbd be equal to the sum
of the two inward opposite angles a
and c.

For, conceive be to be drawn pa» "^^ x» xr

rallel to the side ao of the triangle.
Then uc, meeting the two parallels ao, be, makes the alter*
nate angles c and cbb equal (th. 12). And ad, cutting the
same two parallels ac, be, makes the inward and outward
angles on the same side, a and sbd, e^al to each other
(th. 14). Therefore, by equal additions, the sum of the two
angles a and c, is equal to the sum of the two cbb and bbs*
that is, to the whole angle cbd (by ax. 3}. t^ x. d.



theorbm xni.

In any triangle, the sum of all the three axgles is equal lo
two right angles.

Let ABC be any plane triangle ; then
the sum of the three angles a + b + c
is equal to two right angles.

For, let the side ab be produced to d.
Then the outward angle cbd is equal
to the sum of the two inward opposite
angles a + c (th. 16). To each of these equals add the in-
ward angle b, then will the sum of the three inward angles
^ ^ B + c be equal to the sum of the two adjacent angles
ABU + cbd (ax. 2). But the sum of these two last adjacent
angles is equal to two right angles (th. 6). Therefore also
the sum of the three angles of the triangle a + b -{- c is
equal to two right angles (ax. 1). a. s. d.

Cord. 1. If two angles in one triangle, be equal to two
angles in another triangle, the third angles will also be equal
(ax. 3], and the two triangles equiangular.

Cord. 2. If one angle in one triangle, be equal lo ette
angle in another, the sums of the remaining angles wiU ak»
be equal (ax. 3).




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281

Cmtl* S» If one angle of r triangle be right, the sum of
the other two will also be equal to a right angle, and each of
them singly will be acute, or less than a right angle.

C&nl. 4. The two least anules of every triangle are acute,
or each less than a right anjrie.

THEOllKM XVIII.

In any quadrangle, the sum of all the four inward angles, is
equal to four right angles.
Let ABCD be a quadrangle ; then the
ram of the four inward angles, a + » +
c 4" i> is equal to four right angles.

Let the diagonal ac be drawn, dividing
the quadrangle into two triangles, abc, ado.
Tlien, because the sum of the three angles
of each of these triangles is equal to two
right angles (th. 17) ; it follows, that the sum of all the
angles of both triangles, which make up the four angles of
the quadraiigle, must be equal to four right angles (ax. *^).

a. E. D.

Carol, 1. Hence, if three of the angles be right ones, tie
fourth will also be a right angle.

Cani. 2. And if the sum of two of the four angles be
equal to two right angles, the sum of the remaining two wiH
«lso be equal to two right angles.

THEOKEK XrX.

Iif any figure whatever, the sum of all the inward angles,
taken tosether, is equal to twice as many right angles,
wanting four, as the figure has sides.

Let ABODE be any figure ; then the
aum of all its inward angles, a + b +
c + D + B» ^'' equal to twice as many
light angles, wanting four, as the figure
kas sides.

For, from any point p, witliin it, draw
lines, PA, PB, PC, dec. to all the angles,
dividing the polygon into as many tri-
angles as it has Hides. Now the sum of the three angles of
each of these triangles, is equal to (wo ri^ht angles (th. 17) ;
Iberefore the sum of the angles of ail the triangles is equii
to twice as many right angles as the figure has sides. But
tbe iOffl of all the angles about the point p, which aj-e ao




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392



OBOKSTIKY.



many of tlie angles of the triangles, but no part of tlie ia-
ward angles of the polygon, is equal to four right angles
(corol. 3, th. 6), and must be deducted out of the former
sum. Hence it follows that the sum of all the inward angles
of the polygon alone,. A + b.+ c + n + ', «s equal to twice
as many right angles as the figure has sides, wanting the
said four right angles, a. e. d.




THEOREM XX,

When every side of any figure is produced out^ the sum
of all the outward angles thereby made, is equal to four right
angles.

' Let A, B, c, dec. be the outward
angles of any polygon, made by pro-
ducing all the sides ; then will the sum
A + B + c + i> + E> of all those outward
angles, be equal to four right angles.

For every one of these outward an-
gles, together with its adjacent inward
angle, make iTp two right angles, as
A+a equal to two right angles, being
the two angles made by one line meeting another (th. 6).
And there being as many outward, or inward angles, as the
figure ' has sides ; therefore the sum of all the inward aad
outward angles, is equal to twice as many right angles as
the figur^ has sides. But the sum of all the inward angles
with four right angles, is equal to twice as many right angles
as the figure has sides (th. 19). Therefore the sum of all
the inward and all the outward angles, is equal to the sum
of all the inward angles and four right angles (by ax. 1).
From each of these take away all the inward angles, and
there remains all the outward angles equal to four right angles
(by ax. 3).



THEOREH XXI.

A pevfendicular is the shortest line that can be drawn
from a given point to an indefinite line. And, of any other
linos drawn from the same point, those that are nearest the
perpendicular are less than those more remote.

If AB, AC, AD, d^. be lines drawn from
the given point a, to the indefinite line de,
of which AB* is perpendicular ; then shall
the perpendicular ab be less than ac, and
AC less than ad, dec.

Fori the angle b being a right one» the



A



D C B £



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anglf o 18 acQte (by cor. 8, th. 17), and therefore less than
the angle b. But the less angle of a triangle is subtended
by the less side (th. 9). Therefore the side ab is less than
the side ac.

Again, the angle acb being acute, as before, the adjacent
angle acd will be obtuse (by th. 6) ; consequently the angle
D is acute (corol. 3, th. 17), and therefore is less than the
angle c. And since the less side is opposite to the less angle,
therefore the side ac is less than the side ad* o. s. d.

Ccfol. A perpendicular is the least distance of a given
point from a line.



THEOREM XZn.

The opppsite sides and angles of any parallelogram are
equal to each other ; and the diagonal divides it into two
equal triangles.

Let ABCD be a parallelogram, of which
the diagonal is bd ; then will its opposite
sides and angles be equal to each other,
and the diagonal bd will divide it into two






equal parts, or triangles. ^ jg

For, since the sides ab and dg are pa-
rallel, as also the sides ad and bc (de6n.
8:2), and the line bd meets them ; therefore the alternate
angles are equal (th. 12), namely, the angle abd to the angle
cDB, and the angle adb to the angle cbd. Hence the two
triangles, having two angles in the one equal to two angles
in the other, have also their third angles equal (cor. 1, th. 17),
namely, the angle a equal to the angle c, which are two of
the opposite angles of the parallelogram.

Also, if to the equal angles abd, cdb, be added the equal
angles cbd, abd, the wholes will be equal (ax. 2), namely,
the whole angle abc to the whole ado, which are the other
two opposite angles of the parallelogram. q. b. d.

Again, since the two triangles are mutually equiangular
and have a side in each equal, vix. the common side bd ;
therefore the two triangles are identical (th. 2), or equal in
all respects, namely, the side ab equal to the opposite side
DC, and AD equal to the opposite side bc, and the whole
triangle abd equal to the whole, tiianf^e dcd. a. b. d.



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394 OBOXBTRT.

Corol. 1. Hence, if one angle of a paralielogram be a ri|^
angle, all the other three wili also be right anglea, and the
parallelogram a rectangle.

Corel. 2. Hence also, the sum of any two adjacent angles
of a parallelogram m equal to two right angles.

TUBORBX XXIII.

EvBRv qnadrilateral, whose opposite sides are equal, is a
parallelogram, or has its opposite sides parallel.

Let AacD be a quadrangle, having the
opposite sides equal, namely, the side ab
equal to dc, and ad equal to bc ; then
shall these equal sides be also parallel,
and the figure a parallelogram.

For, let the diagonal bd be drawn.
Then, the triangles, abd, cbd, being
mutually equilateral (by hyp.), they are
also mutually equiangular (th. 5), or have their co
angles equal ; consequently the opposite sides are parallel
(th. IS) ; viz. the side ab parallel to do, and ad parallel to
BC, and the figure is a parallelogram, a. k. d.

THEORBM XXIV.

TnosB lines which join the corresponding extremes of
two equal and parallel lines, are themselves equal and
parallel.

Let AB, DC, be two equal and parallel lines ; then will
the linei^ ad, bc, which join their extremes, be also equal
and parallel. [See the fig. above.]

For, draw the diagonal bd. Then, because ab and dc are
parallel (by hyp.), the angle abd is equal to the altemato
angle bdc (th. 12). Hence then, the two triangles having
two sides and the contained angles equal, viz. the side ab
equal to the side do, and the side bd common, and the con-
tained angle abd equal to the contained angle bdc, they
have the remaining sides and angles also respectively equal
(th. 1) ; consequently ad is equal to bc, and also parallel to
it (th. 12). a. B. D.

THEORBM XKV.

Parallbloobaxs, as also triangles, standing on the
same base, and between the same parallels, are equal te
each otben



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TKXOBBm. 20S

Let ABCD, ABSF, be two panllelograoiSy ID_
Asc, ABF, two triangles, standing on '^
the same base ab, and between the same
pafailels ab, db ; then will the paraHelo-
gram abcd be equal to the parallelogram
ABBF) and the triangle abc equal to the
triangle abf«

For, since the line dk cuts the two parallels af, bb, and
the two AD, BC, it makes the angle b equal to the angle afd,
and the angle d equal to the angle bob (th. 14) ; the two
triangles adf, bce, are therefore equiangular (cor. 1, th. 17) ;
and having the two corresponding sides ad, bc, equal
(th. 23), being opposite sides of a parallelogram, these two
triangles are identical, or equal in all respects (th. 2). If
each of these equal triangles then be taken from the whole
apace abbd, there will remain the parallelogram abef in
the one case, equal to the parallelograms abcd m the other
(by az. 3).

Also the triangles abc, abf, on the same base ab, and
between the same parallels, are equal, being the halves of
the said equal parallelograms (th. 22). o. b. d.

CardL !• Parallelograms, or triangles, having the same
base and altitude, are equal. For the altitude is Ste same as
the perpendicular or distance between the two parallels, which
is every where equal, by the definition of parallels.

Carol, 2. Parallelograms, or triangles, having equal bases
and altitudes, are equal. For, if the one figure be applied
with its base on the other, the bases will coincide or be the
same, because they are equid : and so the two figures* having
the same base and altitude, are equal.

THEORBX XXVI.

If a parallelogram and a triangle, stand on the same
base, and between the same parallels, the parallelogram
will be double the triangle, or the triangle half the paral-
lelogram.

Let abcd be the parallelogram, and abb a
triangle, on the same base ab, and between
the same parallels ab, db ; then will the
parallelogram abcd be double the triangle
ABE, or the triangle half the parallelo-
gram.

For, draw the diagonal ac of the parallelogram, dividiBg
it into two equal parts (th. 22). Then because the Irianglee




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396, esoxBTBT.

AMOp ABB, on the same base, and between the same parallelsr
are equal (th. 25) ; and because the one triangle abc is half
the parallelogram abcd ^th. 22), the other equal triangle
ABB is also equal to half the same parallelogram abcd.
a. B. B.

CoroL I. A triangle is equal to half a parallelogram of the
same base and altitude, because the altitude is the perpendi-
cular distance between the parallels, which is every where
equal, by the definition of parallels.

Card. 2. If the base of a parallelogram be half that of a
triangle, of the same altitude, or the base of the triangle be
double that of the parallelogram, the two figures will be
equal to each other.

THBOBEX ZXVn.

Rbctaboxbs that are contained by equal lines, are equal
to each other.

Let bd, fb, be two rectangles, having p C H O
the sides ab, bc, equal to the sides kf,
FO, each to each ; then will the rectangle
BD be equal to the rectangle fh,



For, draw the two diagonals ac, eo, ATB — -
dividing the two parallelograms each into
two equal parts. Then the two triangles abc, efg, are
equal to each other (th. 1), because they have the two sides
AB, BO, and the contained angle b, equal to the two sides
XF, FG, and the contained angle f (by hyp.). But these
equal triangles are the halves of the respective rectangles.
And because the halves, or the triangles, are equal, the
wholes, or the rectangles db, hf, are abo equal (by az. 6).

Q. B. D.

Cord. The squares oh equal lines are also equal; for
every square is a species of rectangle.

THBOBEX XXVin.

Thb complements of the parallelograms, which are about
the diagonal of any parallelogram, are equal to each other.

Let AC be a parallelogram, bv a dia-
gonal, aiF parallel to ab or dc, and oih
parallel to ad or bc, making ai, ic, com-
plements to the parallelograms bo, hf,

which are about the diagonal db : then ^ ^

will the complement az be equal to the /^ jj

complement ic.




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39T

/

For, tinee the Aagonftl db bisects the three parallelogranw
jLCf B0| HF (th. 22) ; therefore, the whole triangle dab being
eqaal to the whole triangle dcb, and the parts dbi, irb, re-
spectively equal to the parts doi, ifb, the remaining parts Ar,
10, must also be equal (by ax,. 3). q. b. d.

THEORBX XXIX.

A TRAPBZOiD, or trapezium having two sides parallel, is '
equal to half a parallelogram, whose baqe is the sum of those
two sides, and its altitude the perpendicular distance between
them.

Let ABCD be the trapezoid, having its d r TT V*

two sides ab, Dc, parallel ; and in ab i^ i- H F

pioduced take be equal to dc, so that
AS may be the sum of the two parallel



f



EJ



sides $ produce dc also, and let bf, oci A. 6 B £
BH» be all three parallel to ad. Then is
AF a parallelogram of the same altitude with the trapezoid
ABCD, having its base ab equal to the sum of the parallel
aides of the trapezoid ; and it is to be proved that the trape-
zoid ABCD is equal to half the parallelogram af.

Now, since triangles, or parallelograms, of equal bases and
altitude, are e<pial (corol. 2, th. 25), the parallelogram do is
equal to the parallelogram he, and the triangle cob equal to
the triangle chb ; consequently the line bc bisects, or equal-
ly divides, the parallelogram af, and abcd is the half of it.
a« B. D.

thbobbx txz.

TsB sum of all the rectangles contained under one whole
line* and the several parts of another line, any way divid*
ed, is equal to the rectiemgle contained under the two whole
lines.

Let AD be the one line, and ab the r jr f*

other, divided into the parts ab, bf, " P . V /

FB ; then will the rectangle contained
by AD and ab, be equal to the sum of
the rectangles of ad and ab, and ad and



■Fy and AD and fb : thus expressed, -^ E P B

AD • ab := AD • ab + AD . SF + AD • FB.

For, make the rectangle ac of the two whole luies ad,
ab ; and draw bo, fb, perpendicular to ab, or parallel to
AD, to which they are equal (th. 22). Then the whole
rectangle ac is made up of all the other rectangles ao, eh.

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298



OEOMEnr.



6 H C



^ ic. But these rectangles are eontaia^
ed by as and ab, bo and ef^ fh and fb ;
which are equal to the rectangles of ad
and AB, AD and bf, ad and fb, because
AD is equal to each of the two bo, fh.
Therefore the rectangle ad. ab is equal
to the sum of all the other rectangles ad •

AB, AD • EF, AD . FB. Q* E. D.

Card. If a right line be divided into any two parts, the
square on the whole line, is equal to both the rectangles of
the whole line and each of the parts.



A X FB



\



THEOREX XXXI.

The square of the sum of two lines, is greater than the
sum of their squares, by twice the rectangle of the said
lines. Or, the square of a whole line, is equal to the
squares of its two parte, together with twice the rectangle of
those parts.

Let the line ab be the sum of any two
lines AC, cb ; then will the square of ab
be equal to the squares of ac, cb, together
with twice the rectangle of ac . cb. That

is, AB*= AC* + cb" + 2aC . CB.



H T>



G



C B



For, let ABDE be the square on the sum
or whole line ab, and acfg the square
on the part ac Produce cf and gf to the other sides at h
and I.

From the lines ch, oi,'' which are equal, beinff each equal
to the sides of the square ab or bd (th. 22), tf^e the parts
CF, GF, which jare also equal, being the sides of the square
af, and there remains fh equal to fi, which are also equal
to.DH, Di, being the opposite sides of the parallelogram.
Hence the figure hi is equilateral : and it has all its angles
right ones (corol. 1, th. 22) ; it is therefore a square on the
line FI, or the square of ite equal cb. Also the figures BFy
fb, are equal to two rectangles under ac and gb, because
GF is equal to ac, and fh or fi equal to cb. But the
whole square ad is made up of the four figures, viz* the twe
squares af, fd, and the two equal rectangles bf, fb« That
is, the square of ab is equal to the squares of ac^ gb, toge-
ther with twice the rectangle of ac, cb. . a* e« d.

Cord* Hence, if a line be divided into two equal parts ;
the square of the whole line will be equal to four times the
square of half the line.



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THBOBBXfl.




k: I



TBBOKBM XXXn.

Thb square of the difference of two lines, is less than the
mim of their squares, by twice the rectangle of the said
lines.

Let AC, Bc, be any two lines, and ab
(heir difference : then will the square of ab
be leas than the squares of ac, bc, by
twice the rectangle of ac and bc. Or,

ab* s= AC* + Bc' — ^2aC • BC.

For, let ABDB be the square on the'dif.
ference ab, and acfg the square on the
line AC« Produce ed to h ; also produce
OB and HC, and draw ki, making bi the square of the other
linoBC.

Now it is visible that the square ad is less than the two
•qoares af, bi, by the two rectangles ep, di. But gf is
equal to the one line ac, and gb or fr is equal to the other
line BC ;NConsequently the rectangle ef, contained under eg
and OF, is^ual to the rectangle of ac and bc.

Agaio, nf being equal to ci or bc or dh, by adding the
tioinmoB parTHC, the whole hi will be equal to the whole fc,
or equal to ac ; and consequently the figure i>i is equal to
die rectangle contained by ao and bc.

Hence the (wo figures bf, di, are two rectangles of the
two lines ac, bo ; and consequently the square of ab is
teas than the squares ef ac, bc, by twice the rectangle
AC • BC. a* B* D.

theobbk xxxiii«

The reclaaffle under the sum and difference of two lines, is
equal to the difference of the squares of those lines'^.
Let AB, AC, be any two unequal lines ;

tten will the difference of the squares of

AB, AC, be equal to a rectangle under

4faeir sum and difference That is,

ab' — AC^ » AB 4- AC • AB — AC.

For, let ABDB be the square of ab, and
ACFO the square of ac. Produce db
fill BH be equal to ac ; draw hi parallel
to AB or BD, and produce fc both ways
to I and K.



K D



G



A C



B



* Thb and the two preceding theorems, are evinced algebraically,
bf the three expressions
<a-f &)«:=:•« 4.8a6 + 6s = iil + 6s + 9a6
{• — 6)« = 4i?— 8a* + *a = ai + 6a — 2aA
(a + 6)Ca-6) = a^-.ft».



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ia»



Then the difference of tUe two squares ad, ap, is oTf.
dently the two rectangles sf, kb. But the rectangles bf,
Bi are equal, being contained under equal lines ; for ek and
BH are each equal to ac, and oe is equal to cb, being each
equal to the difference between ab and ac, or their equals
AB and AG. Therefore the two ef, kb, are equal to the two
KB, Bf, or to the whole kb ; and consequently kh is equal
to the difference of the squares ad, af. But kh is a rect-
angle contained by dh, or the sum of ab and ac, and by kd,
or the difference of ab and ac. Therefore the difference of
the squares of ab, ac, is equal to the rectangle under their
sum and difference, a. e. d.

THEOSBIC XXXnr.




IiT any right angled triangle, the square of the hypo-
thenuse, is equal to the sum of the squares of the other two
sides.

Let ABC be a right-angled triangle,
having the right angle o ; then will the
square of the hypothenuse ab, be equal
to the sum of the squares of the other ^
two sides ac, cb. Or ab' = ac"

+ B0«.

For, on ab describe the square ae,
and on ac, cb, the squares ao, bh;
then draw ck parallel to ad or be ; rj -

and join ai, bf, cd, cb. D aL S

Now, because the line ac meets the two cg, cb, so as to
make two right angles, these two form one straight line on
(eorol. 1, th. 6). And because the angle fac ia equal to the
angle dab, being each a right angle, or the angle of a square ;
to each of these equals add the common angle bac, so will
the whole angle or sum fab, be equal to the whole an^e or
sum cad. But the line fa is equal to the line ac, and the
line AB to the line ad, being sides of the same square ; so
that the two sides fa, ab, and their included angle fab, are
equal to the two sides ca, ad, and the contained angle cad,
each to each : therefore the whole triangle afb is equal to
the whole triangle acd (th. 1).

But the square ao is double the triangle afb, on the
same base fa, and between the same parallels fa, ob
(th. 26) ; in like manner the parallelogram ak is double the
triangle acd, on the same base ad, and between the same
parallels ad, ck. And since the doubles of equal things,
are equal (by ax. 6} ; therefore the square ao is equal to we
parallelogram ak.



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aoi

In Uke maiinery the other aqaaie bh is proved equal to
the other parallelogram bk. Consequeatly the two aquares
▲G and BB together, are. equal to the two parallelograms ak
and BK together, or to the whole square ak. That is, the
sum of the two squares on the two less sides, is equal to the
square on the greatest side. a. b. n.

Corol. 1. Hence, the squareof either of the two less sides,
is equal to the difference of the squares of the hypothenuse
and the other side (ax. 3) ; or, equal to the rectangle con-
tained by the sum and difference of the said hypothenuse
and other side (th. 33).

CaroL 8. Hence also, if two light-angled trianeles have
two sides of the one equal to two corresponding sides of the
odier ; their thiid sides will also be equal, and the triangles
identical.



THEORSX XXZV.

In any triangle, the difference of the squares of the
two sides, is equal to the difference of the squares of the
segments of the base, or of Uie two lines, or distances,
included between the extremes of the base and the perpen-
dicular.



J A



Let ABC be any triangle, having
CD perpendicular to ab ; then will
the difference of the squares of ac,
Bc, be equal to the difference of
the squares of ad, bd ; that is, ^ *^ ^ * ^w

AC«-BC» = AD«— BD». A BDA DB

For, since ac* is equal to ad* + cd* J /k ♦u o^\

and Bc« is equal to bd« + cd« $ ^^^ "*• ^\^
Theref. the difference between ac* and bc*,
is equal to the difference between ad* + cd*

and BD* + CD*,



Online LibraryCharles HuttonA course of mathematics for the use of academies, as well as private tuition .. → online text (page 22 of 50)