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Book I.

late.

T

PROPOSITION I,

PROBLEM.

O defcribe an equilateral triangle upon a giver finite ftraight line.

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

From the centre A, at

the distance AB, de

a 3. Poftu- fcribe the circle BCD,
and from the centre B, at
the distance BA, describe
the circle ACE; and D
from the point C, in
which the circles cut one
another, draw the straight

b1. Poft. lines CA, CB to the

nition.

points A, B: ABC fhall

be an equilateral triangle.

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Because the point A is the centre of the circle BCD, AC

C

c 11. Defi 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 d 1. Axi- equal to one anotherd; 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 defcribed upon the given ftraight line AB. Which was required to be done.

om.

PROP. II. PROB.

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

FRO

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

From

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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 f 5. Ax. 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.

FROM

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 ftraight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C, the lefs.

A

a 2. I.

E

B

b 3. Poft.

From the point A draw a the ftraight line AD equal to C; and from the centre A, and at the diftance AD, defcribe b the circle DEF; and becaufe A is the centre of the circle DEF, AE is equal to AD; but the ftraight line C is likewife equal to AD; whence A E and C are each of them equal to AD; wherefore the straight line AE is equal to C, and from As the greater of two ». 1. Byftraight lines, a part AE has been cut off equal to C the! Which was to be done.

PRO

Book I

N.

IF

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F two triangles have two fides of the one equal to two fides of the other, each to each; and have likewife the angles contained by those fides equal to one another; their bafes, or third fides, shall be equal; and the two triangles fhall be equal; and their other angles fhall be equal, each to each, viz. thofe to which the equal fides are oppofite.

Let ABC, DEF be two triangles which have the two fides AB, AC equal to the two fides DE, DF, each to each, viz. AB to DE, and AC

to DF; and the angle
BAC equal to the
angle EDF, the bafe
BC fhall be equal to
the bafe EF; and the
triangle ABC to the

A

D

triangle DEF; and

the other angles, to B

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which the equal fides are oppofite, fhall be equal, each to each, viz. the angle ABC to the angle DEF, and the angle ACB to DFE.

For, if the triangle ABC be applied to the triangle DEF, fo that the point A may be on D, and the ftraight line AB upon DE; the point B fhall coincide with the point E, becaufe AB is equal to DE; and AB coinciding with DE, AC fhall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore also the point C fhall coincide with the point F, because AC is equal to DF: But the point B coincides with the point E; wherefore the bafe BC fhall coa cor. def. 3.incide with the bafe E, and shall be equal to it. Therefore alfo the whole triangle ABC fhall coincide with the whole triangle DEF, and be equal to it: and the remaining angles of the one fhall 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 the angle DFE. Therefore if two triangles have two fides of the one equal to two fides of the other, each to each, and have likewife the angles contained by thofe fides equal to one another; their bafes fhall be equal, and

the

the triangles fhall be equal, and their other angles, to which Book I. the equal fides are oppofite, fhall be equal, each to each. Which was to be demonftrated.

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TH

HE angles at the bafe of an Ifofceles triangle N. are equal to one another; and, if the equal fides be produced, the angles upon the other fidé of - the bafe fhall alfo be equal.

Let ABC be an ifofceles triangle, of which the fide AB is equal to AC, and let the ftraight lines AB, AC be produced to D and E, the angle ABC fhall 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 AG equal to AF, the less, and join FC, GB.

a

A

Becaufe AF is equal to AG, and AB to AC, the two fides 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 bafe FC is equal to the base GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one are equal to the remaining angles of the other, each to each, to which the equal fides are oppofite, viz. the angle ACF to the angle ABG, and the angle AFC to the angle AGB: And

D

T

B

G

a 3. I.

b 4.1.

because the whole AF is equal to the whole AG, and the part AB to the part AC; the remainder BF fhall be equal to the remainder CG; and FC was proved to be equal c 3. Ax. to GB, therefore the two fides BF, FC are equal to the two CG, GB, each to each; but the angle BFC is equal to the angle CGB; wherefore the triangles BFC, CGB are equal, and their remaining angles are equal, to which the equal fides are oppofite; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. Now, fince it has been demonftrated, that the whole angle ABG is equal to the

whole

Book I whole ACF, and the part CBG to the part BCF, 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 alfo been proved, that the angle FBC is equal to the angle GCB, which are the angles upon the other fide of the base. Therefore, the angles at the base, &c. Q. E. D.

COROLLARY. Hence every equilateral triangle is alfo equiangular.

a 3. 1.

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F two angles of a triangle be equal to one another, the fides which fubtend, or are oppofite to, those angles, fhall alfo be equal to one another.

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Let ABC be a triangle having the angle ABC equal to the angle ACB; the fide AB is alfo equal to the fide AC.

D

For, if AB be not equal to AC, one of them is greater than the other: Let AB be the greater, and from it cut off DB equal to AC, the lefs, and join DC; therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two fides DB, BC are equal to the two AC, CB, each to each; but the angle DBC is alfo equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC is equal to the b 4. 1. triangle ACB, the lefs to the greater; which is abfurd. Therefore, AB is not

b

unequal to AC, that is, it is equal to it. B
Wherefore, if two angles, &c. Q.E.D.

COR. Hence every equiangular triangle is alfo equilateral.

PROP.

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