Abbildungen der Seite
PDF
EPUB

2

a Io. Def.

[ocr errors]
[ocr errors]

line, the angle CBE is equal

E

Book. T.
to the angle EBA ; in the same
manner, because ABD is a
straight line, the angle DBE is
equal to the angle EBA ; where.
fore the angle DBE is equal to
the angle CBE, the less to the

D
greater; which is impossible ;
therefore two straight lines can- A

B
not have a common segment.

PRO P. XII. PROB.
10 draw a straight line perpendicular to a given
straight line of an unlimited length,

from a given
point without it.

Let AB be the given straight line, which may be produced
to any length both ways, and let C be a point without it. It
is required to draw a straight
line perpendicular to AB from

C
the point C.
Take any point D upon the

E
other side of AB, and from
the centre C, at the distance
CD, describe the circle EGF

HI
meeting AB in F, G; and bi. A F

G
fect FG in H, and join CF,
CH, CG; the straight line CH, drawn from the given point C,
is perpendicular to the given straight line AB.

Because FH is equal to HG, and HC common to the two
triangles FHC, GHC, the two fides FH, HC are equal to the
two GH, HC, each to each ; and the base CF is equal d to the d 's. Def.
base CG; therefore the angle CHF is equal to the angle CHG;
and they are adjacent angles; but when a straight line standing C,8.1,
on a straight line makes the adjacent angles equal to one ano-
ther, each of them is a right angle, and the straight line which
stands upon the other is called a perpendicular to it; therefore
from the given point C a perpendicular CH has been drawn to
the given straight line AB. Which was to be done,

PRO P. XIII, THE O R.
HE angles which one straight line makes with an-

other upon one side of it, are either two right
angles, or are together equal to two riglit angles.

Les

3. Port,

B

e IO. I.

1.

TH

B 4

Book I.

Let the straight line AB make with CD, upon one fide of it, the angles CBA, ABD; these are either two right angles, or are together equal to two right angles. For, if the angle CBA be equal to ABD, each of them is a

E

А.

b II. I.

C 2. Ax.

di. Ax.

D
B

D

B a Def. 1o. right angle; but, if not, from the point B draw BE at right

angles o to CD; therefore the angles CBE, EBD are two right
angleza; and because CBE is equal to the two angles CBA, ABE
together, add the angle EBD to each of these equals; there.
fore the angles CBE, EBD are equal to the three angles CBA,
ABE, EBD. Again, because the angle DBA is equal to the
two angles DBE, EBA, add to these equals the angle ABC,
therefore the angles DBA, ABC are equal to the three angles
DBE, EBA, ABC; but the angļes CBE, EBD have been de-
monstrated to be equal to the same three angles; and things
that are equal to the same are equal d to one another ; therefore
the angles CBE, EBD are equal to the angles DBA, ABC; but
CBE, EBD are two right angles; therefore DBA, ABC are
together equal to two right angles. Wherefore, when a straight
line, &c. Q. E. D.

PRO P. XIV. THEOR.
F, at a point in a straight line, two other straight lines,

upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the fame straight line.

At the point B in the straight
line AB, let the two straight lines

A.
BC, BD upon the oppolite sides
of AB, make the adjacent angles
ABC ABD equal together to
two right angles. BD is in the
same straight line with CB.

E
For, if BD be not in the same
{traight line with CB, let BE be C B

[ocr errors]

D

in

[ocr errors]

in the fame straight line with it; therefore, because the straight Book I. line AB makes angles with the straight line CBE, upon one w fide of it, the angles ABC, ABE are together equal to two a 13. 1. right angles; but the angles ABC, ABD are likewise together equal to two right angles ; therefore the angles CBA, ABE are equal to the angles CBA, ABD: Take away the common angle ABC, the remaining angle ABE is equal to the remaining b 3. Ax. angle ABD, the less to the greater, which is impossible ; therefore BE is not in the fame straight line with BC. And, in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD, which therefore is in the same straight line with CB. Wherefore, if at a point, &c. Q. E. D.

b

[blocks in formation]

a 15

IF
F two straight lines cut one another, the vertical, er op
posite, angles shall be equal.
Let the wo straight lines AB, CD cut one another in the
point E, the angle AEC shall be equal to the angle DEB, and
CEB to AED.

Because the straight line AE
makes with CD the angles CEA
AED, these angles are toge-
ther equal to two right angles.
Again, because the straight line
Dť makes with AB the angles A E B
AED, DEB, there also are
together equal to two right

D angles; and CEA, AED have been demonstrated to be equal to two right angles ; wherefore the angles CEA, AED are equal to the angles AED, DEB. Take away the common angle AED, and the remaining angle CEA is equal to the remaining angle DEB. In the same b 3. As, manner it can be demonstrated that the angles CEB, AED are èqual. Therefore, if two straight lines, &c. Q. E. D.

COR. I. From this it is manifeft, that, if two straight lines cut one another, the angles they make at the point where they cut, are together equal to four right angles.

COR. 2. And consequently that all the angles made by any number of lines meeting in one points are together equal to four right angles.

PRO P.

Book 1.

PRO P. XVI.

THEOR.

[ocr errors]

a

IF

one side of a triangle be produced, the exterior

angle is greater than either of the interior opposite angles.

Let ABC be a triangle, and let its fide BC be produced to D,
the exterior angle ACD is greater than either of the interior
opposite angles CBA, BAC.
Bifect . AC in E, join BE

A
and produce it to F,and make
EF equal to BE; join also
FC, and produce AC to G.

Because AE is equal to
EC, and BE to EF; AE, EB

E
are equal to CE, EF, each to
each ; and the angle AEB is
equal to the angle CEF,be-

B
С

D
cause they are opposite ver-
tical angles; therefore the
base AB is equal to the base
CF, and the triangle AEB to

G
the triangle CEF, and the
remaining angles to the remaining angles, each to each, to which
the equal fides are opposite; wherefore the angle BAE is equal
to the angle ECF; but the angle ECD is greater than the angle
ECH; therefore the angle ACD is greater than BAE: In the
same manner, if the side BC be bisected, it may be demonstrated
that the angle BCG, that is “, the angle ACD, is greater than

,
the angle ABC. Therefore, if one fide, &c. Q. E. D.

[ocr errors][merged small][merged small][ocr errors]
[blocks in formation]

A

NY two angles of a triangle are together less than
two right angles.

А
Let ABC be any triangle ;
any two of its angles together
are less than two right angles.

Produce BC to D; and be.
cause ACD is the exterior angle
of the triangle ABC, ACD is
greater a than the interior and
oppofite angle ABC; to each of B

С D

these

2 16. I.

these add the angle ACB; therefore the angles ACD, ACB are Book 1: greater than the angles ABC, ACB; but ACD, ACB are to m gether equal to two right angles; therefore the angles ABC b 13. 1. BCA are less than two right angles. In like manner, it may be demonstrated that BAC, ACB, as also CAB, ABC are less than two right angles. Therefore any two angles, &c. Q. E. D.

PRO P. XVIII. THEOR.

THE greater side of every triangle is opposite to the

greater angle.
Let ABC be a triangle, of
which the fide AC is greater
than the side AB; the angle

A.
ABC is also greater than the
angle BCA.
Because AC is greater than

D
AB, make * AD equal to AB,
and join BD; and because A DB

B is the exterior angle of the triangle BDC, it is greater than

b 16.1. the interior and opposite angle DCB; but ADB is equal to c s. 1, ABD, because the side AB is equal to the side AD, therefore the angle ABD is likewise greater than the angle ACB ; wherefore much more is the angle ABC greater than ACB. Therefore the greater fide, &c. 0. E. D.

a 3. t.

с

PRO P. XIX. THE O R.
THE greater angle of every triangle is fubtended by

the greater side, or has the greater side opposite to it. Let ABC be a triangle, of which the angle ABC is greater than the angle BCA; the side AC is likewise greater than the fide AB.

For, if it be not greater, AC must either be equal to AB, or

A leis than it; it is not equal, because then the angle ABC would be equal to the angle ACB; but it is not; therefore AC is not equal to AB; neither is it less; because then the angle B

a s.

ABC

« ZurückWeiter »