XXIV.

An acute angled triangle, is that which has three acute angles.

XXV.

Of four fided figures, a square is that which has all its fides equal, and all its angles right angles.

XXVI.

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

XXVII.

A rhombus, is that which has all its fides equal, but its angles are not right angles.

XXVIII.

A rhomboid, is that which has its oppofite fides equal to one another, but all its fides are not equal, nor its angles right angles.

XXIX.

All other four fided figures befides thefe, are called Trapeziums.

Book L.

Parallel ftraight lines, are fuch as are in the fame plane, and which, being produced ever fo far both ways, do not meet.

L

POSTULATES.

I.

ET 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.

AXIOM S.
I.

HINGS which are equal to the fame thing are equal to one another.

TH

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 fame, are equal to one another.

VII.

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

VIII.

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

IX.

The whole is greater than its part.

X.

All right angles are equal to one another.

XI.

"Two ftraight lines cannot be drawn through the fame point, "parallel to the fame ftraight line, without coinciding with 66 one another.

Book I.

PROPO

Book I.

late.

T

PROPOSITION I. PROBLEM.

O defcribe an equilateral triangle upon a given finite straight line.

Let AB be the given straight line; it is required to defcribe an equilateral triangle upon it.

From the centre A, at

the diftance AB, de

a 3. Poftu- scribe a the circle BCD, and from the centre B, at the distance BA, defcribe the circle ACE; and from the point C, in which the circles cut one another, draw the straight b. Poft. lines b CA, CB to the points A, B: ABC fhall

finition.

be an equilateral triangle.

Because the point A is the centre of the circle BCD, AC 11. De 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 to the fame are 1. Axi. equal to one another d; 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.

om.

F

PROP. II. PROB.

ROM a given point to draw a straight line equal to a given ftraight line.

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

From

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 f BG: But it has been fhewn, that BC is f3. Az equal to BG; wherefore AL and BC are each of them equal to BG; and things that are equal to the fame are equal to one another; therefore the ftraight line AL is equal to BC. Wherefore from the given point A a ftraight line AL has been drawn equal to the given straight line BC. Which was to be done.

F

PROP. III. PROB.

ROM the greater of two given ftraight lines to
cut off a part equal to the lefs.

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 a the ftraight line AD equal to C; and from the centre A, and at the distance AD, defcribe b the circle DEF; and because A is

A

T

E

B

a 2. z.

b3. Poft.

the centre of the circle DEF, AE fhall be equal to AD; but the ftraight line C is likewife equal to AD; whence AE and C are each of them equal to AD; wherefore the ftraight line AE is equal to C, and from AB, the greater c 1. Ax. of two ftraight lines, a part AE has been cut off equal to C the lefs.

C

Which was to be done.

PROP.