are others besides these, but these are sufficient for the BOOK I. purpose. The person who admits these will find himself compelled to admit all that follows to the end of the First Book of Euclid.

If the reader finds any difficulty in understanding or admitting the truth of the last Axiom, he may omit all further consideration of it till he has read to the end of the 28th Proposition of the First Book.]

BOOK I.

a 3 Postu

late.

1 Post.

PROPOSITION I. PROBLEM.

To describe an equilateral triangle upon a given finite straight line.

Let AB be the given straight line:

it is required to describe an equilateral triangle upon it.

From the centre A, at the

distance AB, describe

the circle BCD;

and from the centre B,

at

the distance BA, describe the circle ACE; and from the point C, in

which the circles cut

B

E

one another, draw the straight lines", CA, CB, to the points AB;

ABC shall be an equilateral triangle.

Because the point A is the centre of the circle BCD,

tion.

d 1 Axiom.

and because the point B is the centre of the circle ACE,
BC is equal to BA.

But it has been proved that AC is equal to AB;

therefore AC, BC are each of them equal to AB;

but things which are equal to the same are equal to one another d

therefore AC is equal to BC;

wherefore AC, AB, BC are equal to one another;

and the triangle ABC is therefore equilateral,
and it is described upon the given straight line AB.
Which was required to be done.

[There is no necessity in drawing this figure to complete the BOOK I. circles. Place one point of the compasses on A and the other

point on B. Then with centre A and distance AB draw a small arc near where the two circles are likely to intersect. Then with centre B and distance BA touch the arc in the point C. Then with a pen and ruler join CA, CB.

A

B

Before proceeding further, practise drawing equilateral triangles as follows:

Having made the triangle ABC, describe another equilateral

triangle on the opposite F

C

D

side of AB, and also one

on BC and on AC. The three triangles thus constructed can be made to lie exactly upon ABC. Cut the figure out in paper, and turn each of the triangles AEB, CDB, AFC on AB, BC, CA

E

as hinges till the points D, E, F coincide. The figure thus formed is the first of the "five regular solids, and is called a tetrahedron."]

PROP. II. PROB.

From a given point to draw a straight line equal to a given straight line.

Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC.

BOOK I. From the point A to B draw the straight line AB; and upon it describe the equila

1 Post.

bi. 1.

C 2 Post.

d 3 Post.

teral triangle DAB,

and produce the straight lines
DA, DB, to E and F ;

from the centre B, at the dis-
tance BC, described the circle
CGH;

and from the centre D, at the dis

tance DG, describe the circle
GKL.

AL shall be equal to BC.

Because the point B is the centre of the circle CGH,

15 Def. BC is equale to BG;

13 Ax.

h 1 Ax.

and because D is the centre of the circle GKL,

DL is equal to DG,

and DA, DB, parts of them, are equal;

therefore the remainder AL is equal to the remainder BG. But it has been shown that BC is equal to BG;

wherefore AL and BC are each of them equal to BG;

and things that are equal to the same are equal to one another h;

therefore the straight line AL is equal to BC.

Wherefore, from the given point A,

a straight line AL has been drawn equal to the given straight

line BC.

Which was to be done.

[For the future, when it is required to draw a right line from a given point A equal to a

given right line, BC, place the
two points of your compasses
on B and C and remove one of

B

A

L

them to A, and the other to L, and with a pen and ruler draw the straight line AL.]

BOOK I.

PROP. III. PROB.

From the greater of two given straight lines to cut off a part

equal to the less.

Let AB and C be the two given straight lines,

whereof AB is the greater.

It is required to cut off from AB,

the greater, a part equal to
C, the less.

From the point A draw the straight line AD equal to C, and from the centre A, and at the distance AD, describe the circle

DEF;

And because A is the centre of the circle DEF,
AE shall be equal to AD;

but the straight line C is likewise equal to AD;
whence AE and C are each of them equal to AD;
wherefore the straight line AE is equal to C,
and from AB, the greater of two straight lines,
a part AE has been cut off equal to C the less.
Which was to be done.

c 1 Ax.

[There is no necessity to draw this figure whenever it is required to cut off a part from the greater of two straight lines equal to the less.

the same distances from each other, place them respectively

on A and E.]