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therefore BCcoincidingwith EF,BAand ACshall coincide with ED Book I. and DF. for if the base BC coincides with the base EF, but the ades BA, CAdo not coincide with the sides ED,FD, but have a different situation, as EG,FG; then upon the same base EFand upon the same 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 likewise their sides terminated in the other extremity. but this is impossible a. therefore if the base BC coincides with the base EF, a. 7. I. the sides BA, AC cannot but coincide with the sides 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. b. 8. A.

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b. I. I.

PROP. IX. PRO B.
10 bisect a given rectilineal angle, that is, to divide

it into two equal angles.
Let BAC be the given rectilineal angle, it is required to bisect it.

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

Because AD is equal to AE, and AF D E is common to the two triangles DAF, EAF; the two sides DA, AF are equal to the two fides EA, AF, each to each; and the base DF is equal to the basé By

F EF; therefore the angle DAF is equal e to the angle EAF. wherefore the given rectilineal angle BAC c. 8. 1. is bisected by the straight line AF. Which was to be done.

TO

PRO P. X. PROB.
10 bisect a given finite straight line, that is, to di-

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 b a. 1. 8.

b. g. 1. the angle ACB by the straight line CD. AB is cut into two equal parts in the point D.

Book I.

Because 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; there-
fore the base AD is equal to the base o
DB, and the straight line AB is divided
into two equal parts in the point D. A
Which was to be done.

C. 4. I.

D

B

TO

See N.

a. 3. I.

b. s. I.

PRO P. XI. PRO B.
O draw a straight line at right angles to a given

straight line, from a given point in the same.
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 make a CE equal to CD, and
upon DE describe b the equilateral
triangle DFE, and join FC. the

F
Itraight 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 gdes AD E B
DC, CF are equal to the two EC,
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

straight line makes with another straight line are equal to one d. 10. Def. another each of them 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 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 demonstrated that two straight lines cannot have a common segment.

If it be possible, 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, the

c. 8. I.

I.

E

angle CBE is equal to the angle

Book I. EBA, in the fame manner, because

a. 10.Def.i. ABD is a straight line, the angle DBE is equal to the angle EBA. wherefore the angle DBE is equal to the angle CBE, the less to the

D
greater; which isimpoflible. there-
fore two straight lines cannot have A B
a common segment.

PROP. XII. PROB.
O 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 the point C.

Take any point D upon the other side of AB, and from the center C, at the distance CD, describe b the circle EGF meeting AB in F, G; and bisect

HI FG in H, and join CF, CH, CG. A F

G B the straight line CH drawn from

D the given point C, is perpendicular to the given straight line AB.

Because FH is equal to HG, and HCcommon to the two triangles FHC, GHC, the two sides FH, HC are equal to the two GH, HC, each to each; and the base CF is equal d to the base CG; therefore d.rs.Def.r. the angle CHF is equal o to the angle CHG; and they are adjacent e. 8. I? angles. but when a straight line standing 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 afperpen- f.10. Def.i. dicular to it. thereforefrom thegiven point Ca perpendicular CH has · been drawn to the given straight line AB. Which was to be done.

PROP. XIII. THEO R.
HE angles which one straight line makes with

another upon one side of it, are either two right angles, or are together equal to two right angles,

b.

Post

C. IO. I.

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Book I.

Let the straight line AB make with CD, upon one side of its the angles CBA, ABD; these are either two right angles, or are

together equal to two right angles. a. Def. 1o. For if the angle CBA be equal to ABD, each of them is a right A

E

A

d. 1. Ax.

De B

D B В b. 11. 1. angle. but if not, from the point B draw BE at right anglešbto 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. 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 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.

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.

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

A
BC, BD upon the opposite fides
of AB, make the adjacent angles
ABC, ABD equal together to
two right angles. BD is in the

E fame straight line with CB.

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

I'

in the fame straight line with it. therefore because the straight line Book 1. AB makes angles with the straight line CBE, upon one side of it, the angles ABC, ABE are together equal a to two right angles; a. 13. I. 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 o to the reinaining angle ABD, the less to the b. 3. A&. 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 same straight line with CB. Wherefore if at a point, &c. Q. E. D.

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

opposite, 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 together equal a to two right angles. again,

a. 13. I. because the straight line DE makes with AB the angles AED, DEB; these also are together equal • to two right angles. and CEA, AED A have been demonstrated to be equal to two right angles; where-, fore the angles CEA, AED are equal to the angles AED, DEB. take away the common angle AED, and the remaining angle CEA is equal o to the remaining b. 3. As. angle DEB. In the same 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. Aud consequently that all the angles made by any number of lines meeting in one point, are together oqual to four right angles.

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