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For, if the triangle ABC be applied to the triangle DEF, so that the point B may be on E, and the straight line BC upon EF, the point C shall also coincide with the point F, because BC is equal to EF (Hyp.): And BC coinciding with EF, BA and AC shall coincide with ED and DF; for, if the base BC coincides with the base EF, whereas the sides BA, CA do not coincide with the sides ED, FD, but have a different situation as EG, FG, then upon the same base EF, and upon the same side of it, there would be two triangles having their sides terminated in one extremity of the base equal to one another, and also those terminated in the other extremitywhich is impossible (1.7): Therefore, if the base BC coincides with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; and therefore also the angle BAC coincides with the angle EDF, and is equal to it.

Wherefore, If two triangles &c.

Q. E. D.

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To 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 AD; join DE, and upon the side opposite to that on which A lies, describe an equilateral triangle DEF, and join AF: the straight line AF bisects the angle BAC.

D

E

BF C

Because AD is equal to AE, and AF is common to the two triangles DAF, EAF, the two sides DA, AF are equal to the two sides EA, AF, each to each, and the base DF is equal to the base EF

C

--therefore (1. 8) the angle DAF is equal to the angle EAF: Wherefore the given rectilineal angle BAC is bisected by the straight line AF. Q. E. F.

PROP. X. PROB.

To bisect a given finite straight line, that is, to divide it into two equal parts.

Let AB be the given straight line: it is required to bisect it.

Upon it describe an equilateral triangle ABC, and bisect the angle ACB by the straight line CD (1.9): the straight line AB is bisected in the point D.

Because AC is equal to CB and CD is common to the two triangles ACD, BCD, the two sides AC, CD are equal to the two BC, CD, each to each, and the angle ACD is equal to the

angle BCD (Constr.)—therefore the base AD A D B is equal to the base DB: Wherefore the straight line AB is bisected in the point D.

Q. E. F.

PROP. XI. PROB.

To draw a straight line at right angles to a given straight line from a given point in the same.

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

Take any point D in AC, and make CE equal to CD

(1.3), and upon DE describe the equi

lateral triangle DFE, and join CF: the straight line CF is drawn from the given point C at right angles to the given straight A line AB.

F

CEB

Because DC is equal CE, and CF common to the two triangles DCF, ECF, the two sides DC, CF are equal to the two EC, CF, each to each, and the base DF is equal to the base EF-therefore (1. 8) the angle DCF is equal to the angle ECF; and they are adjacent angles : But, when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of these angles is called a right angle (Def. 10); therefore each of the angles DCF, ECF is a right angle: Wherefore, from the given point C, in the given straight line AB, CF has been drawn at right angles to AB. Q. E. F. COR. By the 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 segment AB common to both of them. From the point B draw BE at right angles to AB: Then, because ABC is a straight line, the angle CBE is equal to the angle EBA (Def. 10): So also, because ABD is a straight line, the angle DBE is equal to the angle EBA: Therefore the angle DBE is equal to the angle CBE, the less to the greater--which is absurd: Where

B

E

fore two straight lines cannot have a common segment.

PROP. XII. PROB.

To draw a straight line, perpendicular to a given straight line of 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 from the point C a straight line perpendicular to AB.

E

Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe the circle EGF meeting AB in F, G; bisect FG in H (1. 10), and join CF, CH, CG: the straight line CH is drawn from the given point C perpendicular to the given straight line AB.

H

D

GB

Because FH is equal to HG, and HC common 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 to the base CG-therefore (1. 8) the angle CHF is equal to the angle CHG; and they are adjacent angles: But, when a straight line standing on another straight line makes the adjacent angles equal to one another, each of these angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it (Def. 10); therefore CH is perpendicular to AB: Wherefore, from the given point C without it, CH has been drawn perpendicular to the given straight line AB.

Q. E. F.

PROP. XIII. THEOR.

The 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.

A

Let the straight line AB make with CD, upon one side 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 right angle (Def. 10): But if not, from the point B draw BE at right angles to CD (1. 11); then the

E

D

B

C

DB

angles CBE, EBD are two right angles: And, because

the angle CBE is equal to the two angles CBA, ABE, to each of these equals add the angle EBD; therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD (Ax. 2): Again, because the angle DBA is equal to the two angles DBE, EBA, to each of these equals add the angle ABC; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC: But the angles CBE, EBD have been shewn to be equal to the same three angles; and things that are equal to the same 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, The angles, which one straight line &c. Q.E.D.

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If, 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 same straight line.

E

B

D

At the point B in the straight line AB, let the two straight lines BC, BD, upon the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles: BD shall be in the same straight line with BC.

For, if BD be not in the same straight line with BC, let BE be in the same straight line with it: Then because the straight line AB makes with the straight line CBE, upon one side of it, the angles CBA, ABE, these are together equal to two right angles (1. 13): But the angles CBA, ABD are together equal to two

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