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 a upon it an equilateral triangle ABC,

b

and bisect the angle ACB by the straight line CD.
AB is cut into two equal parts in the point D.

с

therefore the base AD is equal to the base © DB,

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

[The most expeditious mode of
bisecting a straight line is as fol-
lows:

Place the point of the compasses
on A, and with an opening larger
than half the line, draw a small arc
at C above, and another at E be- A
low the line. Then with the point
at B and the same aperture touch
the arcs in the point C and E.
Then with a ruler placed upon C
and E, observe where it cuts the
line AB, and mark the point D.]

Di

*

B

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 Cat right angles to AB.

Take any point D in AC,

F

BOOK I.

the straight 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 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;

с

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 another,

d

each of them is called a right 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.

e viii. 1.

a 10 Def.

BOOK I.

[Here again it is not necessary that the triangle DFE be equilateral. It is sufficient if it be isosceles, having DF equal to FE.]

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

right angles to AB; and be

cause ABC is a straight line,

10 Def. 1. the angle CBE is equal to the angle EBA;

in the same manner, because ABD

E

B

wherefore the angle DBE is equal to the 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.

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.

a

describe the circle EGF meeting AB in FG; and bisect FG in H,

and join CF, CH, CG;

the straight 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 sides FH, HC are equal to the two GH, HC, each

to each;

and the base CF is equal to the base CG;

d

therefore the angle CHF is equal to the angle CHG;
and they are adjacent angles; but when a straight line stand-
ing 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
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.

[Instead of describing the arc of a circle, touch the points F and G with the points of the compasses.]

BOOK I.

a 3 Post.

b x. l.

c 15 Def. I.

a viii. 1.

BOOK I.

PROP. XIII. THEOR.

The angles which one straight line makes with another upon the 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 are either two right angles, or are together equal to two right angles.

b xi. l.

c 2 Ax.

d 1 Ax.

but, if not, from the point B

b

draw BE at right angles to CD;

therefore the angles CBE, EBD are two right angles";
and because CBE is equal to

the two angles CBA, ABE together,

add the angle EBD to each of these equals;

therefore the angle 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 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;

d

and things that are equal to the same are equal to one

another;