Page images
PDF
EPUB

XXX.
Of four-sided figures, a square is that

which has all its sides equal, and all
its angles right anglese

XXXI.
An oblong is that which has all its an-

gles right angles, but has not all its
sides equal.

[merged small][ocr errors][merged small][merged small]

XXXIV. 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.

IJI. And that a circle may be described from any

centre, at any distance from that centre.

a

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 remain

ders 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. Tag es 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 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.

66

Real

*

[ocr errors]

PROP. I. PROBLEM.
To describe an equilateral triangle upon a given

finite straight line.
Let AB be the given straight line; it is re-
quired to describe an equilateral triangle upon
AB.

From the centre A, at the *3 Pos. distance AB, describe * the tulate. circle BCD, and from the centre B, at the distance BA, (D

B E describe the circle À CE; 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.

Because the point A is the centre of the circle * 15 De- BCD, AC is equal* to AB; and because the tinition.

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.

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. # 1 Post.

From the point A to B draw* the straight line

AB; and upon it describe* the equilateral tri-. *2 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

iom.

[ocr errors]

* 1. 1.

[ocr errors]

*15 Def.

[ocr errors]

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

K
centre of the circle CGH, BC
is equal* to BG; and because D H
is the centre of the circle GKL, D
DL is equal to DG; and † DA,

Const.

BI DB, parts of them, are equal ; therefore the remainder AL is

F equal to the remainder*BG : 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 t 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.

*3 Ax,

PROP. III. PROB.

[ocr errors]

* 2.1.

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

A E B greater, a part equal to C, the less.

F From the point A draw* the 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 t to AD;

but the straight line C is f15 Def. likewise equal t 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.

13

[ocr errors]

Bon

« PreviousContinue »