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BC is equal to EF; therefore BC coinciding with EF, BA and Book 1. AC fhall coincide with ED and DF; for, if the base BC coincides with the bafe EF, but the fides BA, CA do not coincide with the fides ED, FD, but have a different fituation as EG, FG; then, upon the fame base EF, and upon the fame fide of it, there can be two triangles that have their fides which are terminated in one extremity of the base equal to one another, and likewise their fides terminated in the other extremity: But this is impoffible a; therefore, if the base BC coincides with the a 7. 1. bafe EF, the fides BA, AC cannot but coincide with the fides ED, IF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal b to it. Therefore if two triangles, b s. As. &c. Q. E. D.

PROP. IX. PROB.

O bisect a given rectilineal angle, that is, to divide
it into two equal angles.

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Let BAC be the given rectilineal angle, it is required to bi

fect it.

A

b I. I.

Take any point D in AB, and from AC cut a off AE equal to a 3. 1. AD; join DE, and upon it defcribe b an equilateral triangle DEF; then join AF; the straight line AF bifects the angle BAC.

Because AD is equal to AE, and

AF is common to the two triangles D
DAF, EAF; the two fides DA, AF,

E

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EAF; wherefore the given rectilineal angle BAC is bifected by the straight line AF. Which was to be done.

PROP. X. PROB.

O bifect a given finite ftraight line, that is, to di-
vide it into two equal parts.

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Let AB be the given straight line; it is required to divide it

into two equal parts.

Defcribe a upon it an equilateral triangle ABC, and bifect a 1. 1. the angle ACB by the straight line CD.

equal parts in the point D.

B 3

AB is cut into two bg.

Becaufe

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See N.

a 3. I.

b I. I.

c 8. I.

A

PROP. XI. PROB.

C

B'

O draw a straight line at right angles to a given ftraight line, from a given point in the fame.

Tra

Let AB be a given straight line, and C a point given in it it is required to draw a straight line from the point C at right angles to AB.

Take any point D in AC, and a make CE equal to CD, and upon DE defcribe b the equi

lateral triangle DFE, and join
FC; the ftraight line FC drawn
from the given point C is at
right angles to the given
ftraight line AB.

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A D

F

C

Becaufe DC is equal to CE and FC common to the two triangles DCF, ECF; the two E B fides DC, CF, are equal to the two EC, CF, each to each; and the base DF is equal to the bafe EF; therefore the angle DCF is equal to the angle ECF; and they are adjacent angles. But, when the adjacent angles which one ftraight line makes with another straight line are equal to one another, each of them d 10. Def. is called a right d angle; therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C, in the given ftraight line AB, FC has been drawn at right angles to AB. Which was to be done.

I.

COR. By help of this problem, it may be demonstrated, that two ftraight lines cannot have a common fegment.

If it be poffible, let the two straight lines ABC, ABD have the fegment AB common to both of them. From the point B draw BE at right angles to AB; and because ABC is a straight

line,

line, the angle CBE is equala to
the angle EBA; in the fame
manner, because ABD is a
ftraight 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
greater; which is impoffible;
therefore two straight lines can- A
not have a common segment.

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PROP. XII. PRO B.

E

D

B

draw a straight line perpendicular to a given ftraight line of an unlimited length, from a given point without it.

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

perpendicular to AB from the point C.

Take any point D upon the other fide of AB, and from the centre C, at the diftance CD, describe b the

C

s

H

circle EGF meeting AB in F A

F

G; and bifect, FG in H.

D

Book I.

a 10. Def.

I.

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and join CF, CH, CG; the ftraight 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 bafe CF is equal d to the base CG; therefore the angle CHF is equal e to the angle CHG; and they are adjacent angles; but when a ftraight line standing on a straight line makes the adjacent angles equal to one another, each of them is a right angle; and the ftraight 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. THE OR.

HE angles which one straight line makes with another upon the 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 ftraight 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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a Def. 10. right a angle; but, if not, from the point B draw BE at right b II. 1. angies b to CD; therefore the angles CBE, EBD are two light

angles a; and because CBE is equal to the two angles CBA, ABE together, add the angle EBD to each of thefe equals; there2. 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 thefe 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 demonftrated to be equal to the fame three angles; and things are equal to the fame 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.

I. Ax.

that

I

PROP. XIV. THEOR.

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

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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 a to two right a 13. I. 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 b to the remaining angle b 3. A ABD, the less to the greater, which is impoffible; therefore BE is nct in the fame ftraight line with BC. And, in like manner, it may be demonftrated, that no other can be in the fame ftraight line with it but BD, which therefore is in the fame ftraight line with CB. Wherefore, if at'a point, &c. Q. E. D,

PROP. XV. THEOR.

F two ftraight lines cut one another, the vertical, or

IF obvoi fe, aight lineal the equal.

Let the two ftraight lines AB, CD cut one another in the point E; the angle AEC fhall be equal to the angle DEB, and CEB to AED.

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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 b 3. Az. is equal b to the remaining angle DEB. In the fame manner it can be demonstrated that the angles CEB, AED are equal. Therefore, if two ftraight lines, &c. Q. E. D.

COR. 1. 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 confequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles.

PROP.

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