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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.

XXXII.

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

XXXIII.

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

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.

III. And that a circle may be described from any centre, at ang 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 same side of it, taken together, less than two right angles, these straight lines being continually produoed, shall at length meet upon that side on which are the angles which are less than two right angles.'

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

*(References-- Def. 15; ax. 1; post. 1, 3.)
Let AB be the given straight line.
It is required to describe on AB an equilateral triangle.

B

CONSTRUCTION From the centre A, at the distance AB, describe the circle BCD;

(post. 3) from the centre B, at the distance BA, describe the circle ACE; and from the point C, in which the circles cut one another, draw the

straight lines CA, CB, to the points A, B. (post. 1.) Then ABC shall be an equilateral triangle.

DEMONSTRATION

Because the point A the centre of the circle BCD,

therefore AC is equal to AB; (def. 15) and because the point B is the centre of the circle ACE,

therefore 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 same thing are equal to one another ;

(ax. 1)
therefore CA is equal to CB;
wberefore 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.- PROBLEM. From a given point to draw a straight line equal to a given straight line.

(References— Prop. I. 1 ; post. 1, 2, 3; ax. 1, 3 ; def. 15.)

* To be written out, or repeated to the Teacher.

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

K

н

D

B

E

F

CONSTRUCTION

From the point A to B draw the straight line AB; (post. 1.) upon AB describe the equilateral triangle DAB, (1. 1) and produce the straight lines DA, DB, to E and F; from the centre B, at the distance BC, describe the circle CGH,

(post. 3) and from the centre D, at the distance DG, describe the circle GKL.

Then AL shall be equal BC.

DEMONSTRATION

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

therefore BC is equal to BG ; (def. 15) and because D is the centre of the circle GKL

therefore DL is equal to DG, and DA, DB, parts of them, are equal;

therefore the remainder AL is equal to the remainder BG. (ax. 3.) But it has been shown that BC is equal to BG;

wherefore AL and BC are each of them equal to BG; and things which are equal to the same thing are equal to one another ;

therefore the straight line AL is equal to BC. (ax. 1.) 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. - PROBLEM. From the greater of two given straight lines to cut off a part equal to the less.

(References — Prop. 1. 2; def. 15; post. 3; ax. 1.) Let B and C be the given straight lines, of which AB is greater than C.

It is required to cut off from AB the greater, a part equal to C, the less.

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From the point A draw the straight line AD equal to C; (I. 2)
and from the centre A, at the distance AD, describe the circle DEF.

(post. 3.)
Then AE shall be equal to C.

DEMONSTRATION

Because A is the centre of the circle DEF,

therefore AE is equal to AD; (def. 15) but the straight line C is likewise equal to AD; (constr.)

therefore AE and C are each of them equal to AD;

wherefore the straight line AE is equal to C. (ax. 1.) And therefore 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.

PROP. IV.— THEOREM,

If two triangles have two sides of the one equal to two sides of the other, each to each ; and have likewise the angles contained by these sides equal to each other :

then they shall likewise have their bases, or third sides, equal ; and the two triangles shall be equal ; and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite.

(References — Ax. 8, 10.) Let ABC, DEF, be two triangles, which have the two sides AB, AC, equal to the two sides, DE, DF, each to each, viz. AB to DE, and AC to DF;

and the angle BAC equal to the angle EDF.

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