Page images
PDF
EPUB

gle ABC be applied to DEF, so that the point B 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. Therefore BC coinciding with EF, BA and AC shall coincide with ED and DF: for, if the base BC coincides with the base EF, but 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 can be two triangles that have their sides 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; therefore, if the base BC *7.1. coincides with the base EF, 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 *8 Ax if two triangles, &c. Q. E. D.

*

*

PROP. IX. PROB.

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.

*1.1.

Take any point D in AB, and from AC cut* *3. 1. off AE equal to AD; join DE, and upon it describe an equilateral triangle DEF; then join AF: the straight line AF shall bisect the angle BAC.

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;

*

A

DA E

† Const.

B F

and the base DF, is equal to the base EF; there- + Const. fore the angle DAF is equal to the angle EAF: *8. 1. wherefore the given rectilineal angle BAC is

*1. 1.

*9. 1.

† Const.

bisected by the straight line AF. Which was to be done.

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 divide it into two equal parts.

Describe * upon it an equilateral triangle ABC, and bisect* the angle ACB by the straight line CD. AB shall be cut into two equal parts in the point D.

Because AC is equal to CB, and CD common to the two triangles ACD, BCD; the two sides AC, CD are equal to BC, CD, each to each; and the angle ACD is † Const. equal to the angle BCD; therefore the base AD is equal to the A

* 4.1.

*3.1.

*1. 1.

† Const.

*

D B base DB, and the straight line AB is divided into two equal parts in the point D. Which was to be done.

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 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* CE equal
to CD, and upon DE describe the equilateral
triangle DFE, and join FC. The straight line
FC drawn from the given
point C shall be at right
angles to the given straight
line AB.

Because DC is equal to
CE, and FC common to the
two triangles DCF, ECF; A D
the two sides DC, CF are
equal to the two EC, CF,

F

E B

*

each to each: and the base DF is equal to the † Const. base EF; therefore the angle DCF is equal to *8. 1. 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 another, each of them is called a right* *10 Def. 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 segment AB common to both of them. From the point B draw + BE at +11. 1right angles to AB; and because ABC is a straight line, the angle CBE is

*

equal to the angle EBA; in the same manner, because ABD is a straight line, the angle DBE is equal to the angle EBA: wherefore† the angle DBE is equal to the

A

E

*10 Det.

D

†! Ax.

[blocks in formation]

angle CBE, the less to the greater; which is impossible: therefore two straight lines cannot have a common segment.

PROP. XII. PROB.

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

C

Let AB bé 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

H

B

other side of AB, and from the centre C, at the *3 Post. distance CD, describe* the circle EGF meeting *10. 1. AB in F, G, bisect* FG in H, and join CH. The straight line CH, drawn from the given point C, shall be perpendicular to the given straight line AB.

† Const.

*8. 1.

*

Join CF, CG: and 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 *15 Def. CF is equal* to the base CG; therefore 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 them is a right angle, and the straight line which stands upon the other is called a perpen+10 Def. diculart 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.

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.

Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD: these shall either be two right angles, or shall together be equal to two right angles.

For if the angle CBA be equal to ABD, each of

*10 Def. them is a right*

angle: but if

A

not, from the

point B draw

BE at right

[blocks in formation]

*11 1. angles to CD;

B

D

[blocks in formation]

therefore the

*10 Def. angles CBE, EBD* are two right angles, and

because the angle 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 equal to the three angles CBA, ABE, *2 Ax. EBD. Again, because the angle DBA is equal to the two angles DBE, EBA, add to each of these equals the angle ABC, therefore the angles DBA, ABC are equal to the three †2 Ax. 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 thing are equal* to one another; there- *1 Ax. fore 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 †1 Ax. a straight line, &c. Q. E. D.

[blocks in formation]

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.

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 equal together to two right angles: BD shall be in the C same straight line with CB.

E

B

D

For, if BD be not in the same straight line with CB, let BE be in the same straight line with it: therefore, because the straight line AB makes with the straight line CBE, upon one side of it, the angles ABC, ABE, these angles are together equal to two right angles; but the angles ABC, ABD are likewise together equal+ + Hyp. to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD: †2 Ax.

13. 1.

« PreviousContinue »