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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 produced, shall at length meet upon that side on which are the angles which are less than two right angles.

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 AB.

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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.

Because the point A is the centre of the circle BCD,
therefore AC is equal to AB; (def. 15.)

and because the point B is the centre of the angle 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;

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.

PROPOSITION II. PROBLEM.

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.

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From the point A to B draw the straight line AB; upon AB describe the equilateral triangle DAB, and produce the straight lines DA, DB to E and F;

(post. 1.)
(1. 1.)

(post. 2.)

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 the straight line AL shall be equal to BC.

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; (1. 1.)

therefore the remainder AL is equal to the remainder BG; (ax. 3.) but it has been shewn 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.

PROPOSITION III. PROBLEM.

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, of which AB is the greater.

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

D

E B
F

From the point A draw the straight line AD equal to C; (1. 2.) and from the centre A, at the distance AD, describe the circle DEF. (post. 3.)

Then AE shall be equal to C.

Because A is the centre of the circle DEF,
therefore AE is equal to AD; (def. 15.)
but the straight line C is equal to AD; (constr.)
whence 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.

PROPOSITION 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 those sides equal to each other; 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 cach, viz. those to which the equal sides are opposite.

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 included angle BAC equal to the included angle EDF.

Then shall the base BC be equal to the base EF; and the triangle ABC to the triangle DEF; and the other angles to which the equal sides are opposite shall be equal, each to each, viz. the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

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For, if the triangle ABC be applied to the triangle DEF, so that the point A may be on D, and the straight line AB on DE; because AB is equal to DE,

therefore the point B shall coincide with the point E;

and AB coinciding with DE, because the angle BAC is equal to the angle EDF,

therefore the straight line AC shall fall on DF;
also because AC is equal to DF,

therefore the point C shall coincide with F;
but the point B coincides with the point E;
wherefore the base BC shall coincide with the base EF;
because the point B coinciding with E, and C with F,

if the base BC do not coincide with the base EF, the two straight lines BC and EF would inclose a space, which is impossible. (ax. 10.) Therefore the base BC does coincide with EF, and is equal to it. Wherefore the whole triangle ABC coincides with the whole triangle DEF, and is equal to it;

and the other angles of the one coincide with the remaining angles of the other, and are equal to them,

viz. the angle ABC to the angle DEF,

and the angle ACB to DFE.

Therefore, if two triangles have two sides of the one equal to two sides, &c. Which was to be demonstrated.

PROPOSITION V. THEOREM.

The angles at the base of an isosceles triangle are equal to each other; and if the equal sides be produced, the angles on the other side of the base shall be equal.

Let ABC be an isosceles triangle of which the side AB is equal to AC, and let the equal sides AB, AC be produced to D and E. Then the angle ABC shall be equal to the angle ACB, and the angle DBC to the angle ECE.

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In BD take any point F;

from AE the greater, cut off AG equal to AF the less, (1. 3.) and join FC, GB.

Because AF is equal to AG, (constr.) and AB to AC; (hyp.) the two sides FA, AC, are equal to the two GA, AB, each to each;

and they contain the angle FAG common to the two triangles AFC, AGB; therefore the base FC is equal to the base GB, (1. 4.)

and the triangle AFC is equal to the triangle AGB,

also the remaining angles of the one are equal to the remaining angles of the other, each to each, to which the equal sides are opposite; viz. the angle ACF to the angle ABG,

and the angle AFC to the angle AGB.

And because the whole AF is equal to the whole AG,
of which the parts AB, AC, are equal;

therefore the remainder BF is equal to the remainder CG; (ax. 3.) and FC was proved to be equal to GB;

hence because the two sides BF, FC are equal to the two CG, GB, each to each;

and the angle BFC has been proved to be equal to the angle CGB, also the base BC is common to the two triangles BFC, CĞB;

wherefore these triangles are equal, (1. 4.)

and their remaining angles, each to each, to which the equal sides are opposite;

therefore the angle FBC is equal to the angle GCB,

and the angle BFC to the angle CBG.

And, since it has been demonstrated,

that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are also equal;

therefore the remaining angle ABC is equal to the remaining angle ACB, which are the angles at the base of the triangle ABC;

and it has been proved,

that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore the angles at the base, &c. Q. E.D.

COR. Hence every equilateral triangle is also equiangular.

PROPOSITION VI. THEOREM.

If two angles of a triangle be equal to each other, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another.

Let ABC be a triangle having the angle ABC equal to the angle ACB. Then the side AB shall be equal to the side AC.

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For, if AB be not equal to AC,

one of them is greater than the other.

Let AB be greater than AC;

and at the point B, from BA cut off BD equal to CA the less, (1. 3.) and join DC.

Then, in the triangles DBC, ABC,

because DB is equal to AC, and BC is common to both,

the two sides DB, BC are equal to the two sides AC, CB, each to each;

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