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Book 1.

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 con“ tinually produced, shall at length meet upon that fide on “ which are the angles which are less than two right angles. " See the notes on Prop. 29. of Book I.”

PROPO

PROPOSITION I. PROBLEM.

Book I. O describe an equilateral triangle upon a given fi

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

From the centre A, at the di., ftance AB, defcribe the circle

a. 3. Poftu

late. BCD, and from the centre B, at the distance BA, defcribe the circle ACE; and from the point D

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4 C, in which the circles cut one another, draw the straight linest

b. 2d. Poft, CA, CB to the points A, B; ABC ball be an equilateral triangle.

Because the point A is the centre of the circle BCD, AC is equal to AB ; and because the point B is the centre of the c. 15th Deo circle ACE, BC is equal to BA: But it has been proved that CA finition. is equal to AB; therefore CA, CB are each of them equal to AB ; but things which are equal to the same are equal to one another d; therefore CA is equal to CB; wherefore CA, AB, d. If AxiBC are equal to one another; and the triangle ABC is there. om. fore equilateral, and it is described upon the given straight line AB. Which was required to be done,

PRO P. II. PROB.
"ROM 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

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a. 1, Poft. the straight line AB; and upon it describe the equilateral triangle ;

H DAB, and produce the straight

c... Polt. lines DA, DB, to E and F; from : the centre B, at the distance BC, described the circle CGH, and B

d. 3. Poft. from the centre D, at the distance

E DG, describe the circle GKL. AL

G Aball be equal to BC.

Because

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b. I. !.

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Book I. 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 GKL, 15. Def. DL is equal to DG, and DA, DB, parts of them, are equal; $ 3. Ax. therefore the remainder AL is equal to the remainder f 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 are equal to one another ; therefore the
straight line AL is equal to BC. Wherefore from the given
point A a straight line AL has been drawn equal to the given
Atraight line BC. Which was to be done.

PROP. III. PROB.
ROM the greater of two given straight lines to cut

off a part equal to the less.
Let AB and C be the two gi-
ven 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 straight line AD equal to C;

and from the centre A, and at b. 3. Poft. the distance AD, describe the

circle DEF ; and because 4 is
the centre of the circle DEP, AE shall be equal to AD; but the
straight line C is likewise equal to AD; whence AE and Care

each of them equal to AD; wherefore the straight line AE is ç, 1. Ax. equal to C, and 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. THEORE M.
F two triangles have two fides of the one equal to two

sides of the other, each to each ; and have likewise the angles contained by those fides equal to one another ; they shall likewise have their bases, or third fides, equal; and the two triangles shall be equal; and their other angles shall be equal, each to eachi, viz. those to which the equal sides are opposite.

Lef ABC, DEF be two triangles which have the two Gides AB, AC equal to the two Gdes DE, DF, each to each, viz.

AB

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AB to DE, and AC to DF;

Book 1. and the angle BAC equal A

D to the angle EDF, the base BC shall be equal to the base EF; and the triangle ABC to the triangle DEF ; and the other angles, to which the equal Sdes are opposite, shall be equal each to each, viz. the angle ABCB

CE

F to the angle DEF, and the angle ACB to DFE..

For, if the triangle ABC be applied 10 DEF, so that the point A may be on D, and the straight line AB upon DE; the point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, AC shall coincide with DF, be. cause the angle BAC is equal to the angle EDF; wherefore also the point C shall coincide with the point F, because the straight line AC is equal to DF: But the point B coincides with tbe 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 does not coincide with the base EF, two straight lines would inclose a space, which is impossible. Therefore a 10. Ax. the base BC shall coincide with the base EF, and be equal

Wherefore the whole triangle ABC shall coincide with the whole triangle DEF, and be equal to it ; and the other angles of the cne shall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore, if two tri. angles have two fides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those fides equal to one another, their bases shall likewise be equal, and the triangles be equal, and their other angles to which the equal Gdes are opposite shall be equal, each to each. was to be demonstrated.

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T

HE angles at the base of an Isosceles triangle are

equal to one another; and, if the equal fides be produced, the angles upon the other side of the base thall be equal. Let ABC be an Isosceles triangle, of which the Gide AB is eB

qual

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Book I. qual to AC, and let the straight lines AB, AC be produced to

D and E, the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE.

In BD take any point F, and from AE, the greater, cut off a 3. 1.

AG equal' to AF, the less, and join FC, GB.

Because AF is equal to AG, and AB to AC, the two Gdes FA, AC are equal to the two GA, AB, each to each ; and they contain the angle FAG common to the two triangles AFC,

A
AGB; therefore the base FC is e-
b 4. I.

qualb to the base GB, and the tri-
angle AFC to the triangle AGB;
and the remaining angles of the one
are equal b to the remaining angles
of the other, each to each, to which B

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the equal sides are opposite ; viz.
the angle ACF to the angle ABG,
and the angle AFC to the angle F
AGB: And because the whole AF is
equal to the whole AG, of which the D

E parts A3, AC, are equal; the rec 3. Ax

mainder BF shall be equal to the remainder CG; and FC was proved to be equal to GB; therefore the two fides BF, FC are equal to the two CG, GB, each to each; and the angle BFC is equal to the angle CGB, and the base BC is common to the two triangles BFC, CGB; wherefore the triangles are equal, and their remaining angles, each to each, to which the equal fides are opposite; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG: And, fince it has been demonstrated, that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are allo equal ; the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC : And it has also 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.

COROLLARY. Hence every equilateral triangle is also equiangular.

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F two angles of a triangle be equal to one another,

the sides also which subtend, or are opposite to, the cqual angles, shall be equal to one another.

Let

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