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PROP. XII. PROB.

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

Book. I.

a 10. Def.

I.

the point C.

Take any point D upon the other fide of AB, and from the centre C, at the distance

E

CD, describe the circle EGF

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G

B

D

fect FG in H, and join CF,

e Io. I.

CH, CG; the straight line CH, drawn from the given point C,
is perpendicular to the given straight line AB.

1.

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 dr. Def. bafe 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 another, 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.

THE

PROP. XIII. THEOR.

HE angles which one straight line makes with an-
other upon one fide of it, are either two right

angles, or are together equal to two right angles.

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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

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b II. 1.

dr. Ax.

a Def. ro. right angle; but, if not, from the point B draw BE at right angles b to CD; therefore the angles CBE, EBD are two right angles; and because CBE is equal to the two angles CBA, ABE together, add the angle EBD to each of these equals; therec 2. Ax. 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 angles CBE, EBD have been demonstrated to be equal to the fame three angles; and things that are equal to the fame are equal 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.

PROP. XIV. THEOR.

F, at a point in a straight line, two other straight lines, upon the opposite fides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.

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1

in the same straight line with it; therefore, because the straight Book I. line AB makes angles with the straight line CBE, upon one ~ fide of it, the angles ABC, ABE are together equal to two a 13. 1. right angles; but the angles ABC, ABD are likewife 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 same 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 fame straight line with CB. Wherefore, if at a point, &c. Q. E. D.

I

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F two straight lines cut one another, the vertical, or op.
posite, angles shall be equal.

Let the two 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-C ther equal to two right angles. Again, because the straight line DE makes with AB the angles A

AED, DEB, these also are together equal to two right

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angles; and CEA, AED have

E

a 15. B

D

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 fame b 3. Ax, manner it can be demonstrated that the angles CEB, AED are

equal. Therefore, if two straight lines, &c. Q. E. D.

COR. I. From this it is manifest, 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 point are together equal to four right angles.

PROP.

Book L.

PROP. XVI. THEOR.

IF one fide of a triangle be produced, the exterior angle is greater than either of the interior oppofite 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.

a 10. 1.

6 15. 1.

Bisect AC in E, join BE 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 are equal to CE, EF, each to each; and the angle AEB is equal to the angle CEF, be

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G

the triangle CEF, and the

d 15. 1.

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 FEC; therefore the angle ACD is greater than BAE: In the same manner, if the fide BC be bisected, it may be demonstrated that the angle BCG, that is d, the angle ACD, is greater than the angle ABC. Therefore, if one fide, &c. Q. E. D.

A

PROP. XVII. THEOR.

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

A

16. 1.

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 than the interior and
oppofite angle ABC; to each of B

C

D these erior

fite

D

these add the angle ACB; therefore the angles ACD, ACB are Book 1.
greater than the angles ABC, ACB; but ACD, ACB are to-
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
demonftrated that BAC, ACB, as alfo CAB, ABC are less than
two right angles. Therefore any two angles, &c. Q. E. D.

PROP. XVIII. THEOR.

THE greater fide of every triangle is opposite to the
greater angle.

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angle BDC, it is greater than

b 16. 1.

the interior and opposite angle DCB; but ADB is equal to cs. I

ABD, because the fide AB is equal to the fide AD; there-
fore the angle ABD is likewise greater than the angle ACB;

wherefore much more is the angle ABC greater than ACB.

Therefore the greater side, &c. Q. E. D.

PROP. XIX. THEOR.

THE greater angle of every triangle is fubtended by
the greater fide, or has the greater fide opposite to it.

Let ABC be a triangle, of which the angle ABC is greater
than the angle BCA; the fide AC is likewife greater than the

fide AB.

For, if it be not greater, AC
must either be equal to AB, or
Jess than it; it is not equal, be-
cause 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

ABC

a 5.

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