XXX.

Of four-sided figures, a square is that which has all its sides equal, and all its angles right angles

XXXI.

An oblong is that which has all its angles right angles, but has not all its sides equal.

All other four-sided figures besides these, are called Trapeziums.

XXXV.

Parallel straight lines are such as are in the same plane, and which being produced ever so far both ways, do not meet.

POSTULATES.

I.

LET it be granted, that a straight line may be drawn from any one point to any other point.

II.

That a terminated straight line may be produced to any length in a straight line.

III.

And that a circle may be described from any centre, at any distance from that centre.

AXIOMS.

I.

THINGS which are equal to the same thing, are equal to one another.

II.

If equals be added to equals, the wholes are equal.

III.

If equals be taken from equals, the remainders are equal.

IV.

If equals be added to unequals, the wholes are unequal.

V.

If equals be taken from unequals, the remainders are unequal.

VI.

Things which are double of the same, are equal to one another.

VII.

Things which are halves of the same, are equal to one another.

VIII.

Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.

IX.

The whole is greater than its part.

X.

Two straight lines cannot inclose a space.

XI.

All right angles are equal to one another.

XII.

'If a straight line meets two straight lines, so "as to make the two interior angles on the

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same side of it taken together less than "two right angles, these straight lines being continually produced, shall at length meet upon that side on which are the angles "which are less than two right angles." See the notes on Prop. xxix. of Book 1. Oct. Ed.

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Recal

PROP. 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 AB.

From the centre A, at the

* Pos- distance AB, describe* the tulate. circle BCD, and from the

centre B, at the distance BA, D A BE
describe the circle ACE; and

from the point C, in which
the circles cut one another,

*1 Post. draw the straight lines* CA, CB, to the points A, B; ABC shall be an equilateral triangle.

finition.

Because the point A is the centre of the circle *15 De- BCD, AC is equal* to AB; and because the point B is the centre of the circle ACE, BC is equal to BA: But it has been proved that CA is equal to AB; therefore CA, CB, are each of them equal to AB; but things which are equal *1st Ax- to the same thing are equal to one another; therefore CA is equal to CB: wherefore CA, 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.

iom.

*1 Post.

*1. 1.

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.

From the point A to B draw* the straight line AB; and upon it describe the equilateral tri* Post. angle DAB, and produce* the straight lines DA, DB, to E and F; from the centre B, at the dis*3 Post. tance BC, describe* the circle CGH, and from

the centre D, at the distance DG, describe the
circle GKL, AL shall be equal to BC.
Because the point B is the

centre of the circle CGH, BC
is equal to BG; and because D
is the centre of the circle (KL,
DL is equal to DG; and ↑ DA,
DB, parts of them, are equal;
therefore the remainder ALis
equal to the remainder*BG:

F

H

*15 Def.

† Const.

E

*3 Ax.

but it has been shewn, 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 thing are equal to one another; therefore the straight +1 Ax. 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.

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

E B

*2.1.

straight line AD equal to C; and from the centre A, and at the distance AD, describe the *3 Post. circle DEF: AE shall be equal to C.

Because A is the centre of the circle DEF,

AE is equal to AD; but the straight line C is †15 Def. likewise equal† to AD; whence AE and C are +Const. each of them equal to AD: wherefore the straight line AE is equal to* C, and from AB the greater *1 Ax. of two straight lines, a part AE has been cut off equal to C the less. Which was to be done.

Bon

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