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therefore BC coinciding with EF, BA and ACfhall coincide with ED Book I. and DF. for if the bafe BC coincides with the base 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 same base EF and upon the fame side 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 likewife their fides terminated in the other extremity. but this is impoffible. therefore if the base BC coincides with the base EF, a. 7. 1. the fides BA, AC cannot but coincide with the fides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal to it. therefore if two triangles, &c. Q. E. D.

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PROP. IX. PROB.

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

b. 8. Ax.

Let BAC be the given rectilineal angle, it is required to bifect it.
Take any point D in AB, and from AC cut 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 DAF, EAF; the two fides DA, AF are equal to the two fides EA, AF, each to each;

and the base DF is equal to the bafe B EF; therefore the angle DAF is equal

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D

a

E

b. I. I.

given rectilineal angle BAC c. 8. 1. Which was to be done.

PROB.

10 bisect a given finite ftraight line, that is, to di

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vide it into two equal parts.

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

into two equal parts.

Describe a upon it an equilateral triangle ABC, and bifect ba. 1. I. the angle ACB by the ftraight line CD. AB is cut into two b. g. x. equal parts in the point D.

Book I.

C. 4. I.

See N.

a. 3. I.

b. J. I.

c. 8. I.

Becaufe AC is equal to CB, and CD common to the two triangles ACD, BCD; the two fides AC, CD are equal to BC, CD, each to each; and the angle ACD is equal to the angle BCD; therefore the bafe AD is equal to the base c DB, and the straight line AB is divided into two equal parts in the point D. Which was to be done.

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A

PROP. XI. PROB.

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

Let AB be a given ftraight 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 make a CE equal to CD, and upon DE describe the equilateral

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

Because DC is equal to CE,
and FC common to the two tri-

angles DCF, ECF; the two fides AD

4

DC, CF are equal to the two EC,

c

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C

EB

CF, each to each; and the base DF is equal to the base 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 ftraight line are equal to one d. 10. Def. another each of them is called a right angle; therefore each of

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the angles DCF, ECF is a right angle. wherefore from the given point C in the given straight line AB, FC has been drawn at right angles to AB. Which was to be done.

COR. By help of this Problem it may be demonftrated that two straight lines cannot have a common segment.

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 ftraight line, the

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O draw a ftraight 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 the point C.

Take any point D upon the other fide of AB, and from the center C, at the diftance CD, defcribe the circle EGF meeting AB in F, G; and bifect eFG in H, and join CF, CH, CG. the straight line CH drawn from

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B

C. IO. I.

D

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,

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each to each; and the bafe CF is equal to the base CG; therefore d.15.Def.1. the angle CHF is equal to the angle CHG; and they are adjacent e. s. Ip angles. but when a ftraight line standing on a ftraight 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 f perpen- f. 10. Def. 1. dicular to it. therefore from the given point Ca perpendicular CH has ·

been drawn to the given straight line AB. Which was to be done.

TH

PROP. XIII. THEOR.

HE angles which one straight line makes with another upon one fide of it, are either two right angles, or are together equal to two right angles.

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.

a. Def. 10. For if the angle CBA be equal to ABD, each of them is a right

a

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b. 11. 1. angle. but if not, from the point B draw BE at right angles to CD. therefore the angles CBE, EBD are two right angles a. and because, CBE is equal to the two angles CBA, ABE together; add the angle EBD to each of these equals, therefore the angles CBE, EBD are

c

c. 2. Ax. 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.

d. 1. Ax.

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F at a point in a straight line, two other ftraight 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 straight line.

At the point B in the ftraight line AB, let the two ftraight lines, BC, BD upon the oppofite fides of AB, make the adjacent angles ABC, ABD equal together to two right angles. BD is in the fame ftraight line with CB.

For if BD be not in the fame C ftraight line with CB, let BE be

A

E

B

D

a

in the fame straight line with it. therefore because the ftraight line Book 1. AB makes angles with the straight line CBE, upon one fide of it, the angles ABC, ABE are together equal a to two right angles; a. 13. I. 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 angle ABD, the less to the b. 3. ASi greater, which is impoffible. 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 fame ftraight line with it but BD, which therefore is in the same straight line with CB. Wherefore if at a point, &c. Q. E. D.

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F two ftraight lines cut one another, the vertical, or oppofite, angles fhall be 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. Because the straight line AE makes with CD the angles CEA,

AED, these angles are together equal a to two right angles. again, because the straight line DE makes with AB the angles AED, DEB; thefe alfo are together equal to two right angles. and CEA, AED A

a

E

B.

have been demonstrated to be e

qual to two right angles; where-.

8. 13. I.

fore the angles CEA, AED.are

b

equal to the angles AED, DEB. take away the common angle AED, and the remaining angle CEA is equal to the remaining b. 3. AS. 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 ftraight 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.

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