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"If a straight line meets two straight lines, fo as to make the two "interior angles on the fame fide of it taken together less than "two right angles, these straight lines being continually produc "ed 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,”

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PROPOSITION I. PROBLEM.

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 center A, at the diftance AB describe the circle BCD. and from the center B, at the distance BA describe the circle ACE; and from the point C in which the circles cut one another

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Because the point A is the center of the circle BCD, AC is equal

Book I.

a. 3d Poftulate.

b. ad Poft.

to AB. and because the point B is the center of the circle ACE, c. 15th DeBC is equal to BA. but it has been proved that CA is equal to finition. AB; therefore CA, CB are each of them equal to AB. but things which are equal to the fame are equal to one another; therefore d. 1ft AxiCA 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 straight line AB. Which was required to be done.

PROP. II. PROB.

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 ftraight 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 ftraight line AB; and upon it defcribe the equilateral triangle DAB, and produce the straight lines DA, DB to E and F; from the center B, at the distance BC defcribed the circle CGH, and from the center D, at the distance DG defcribe the circle GKL. AL fhall be equal to

BC.

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a. I Poft.

b. 1. 1.

c. a. Poft.

d. 3. Post.

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Book I. Because the point B is the center of the circle CGH, BC is equal n to BG. and because D is the center of the circle GKL, DL is equal e. 15. Def. to DG, and DA, DB parts of them are equal; therefore the remainder f. 3. Ax. AL is equal to the remainder f BG. but it has been fhewn that BC is 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 straight line AL has been drawn equal to the given straight line BC. Which was to be done.

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PROP. III. PROB.

FROM the greater of two given straight 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.

C

D

A

E B

F

From the point A draw the ftraight line AD equal to C; and from the center A, and at the diftance AD defcribeb the circle DEF. and because A is the center of the circle DEF, AE fhall be equal to AD. but the straight 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 of two straight lines, a part AE has been cut off equal to C the lefs. Which was to be done.

PROP. IV. THEOREM.

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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 thofe fides equal to one another: they fhall likewise have their bases, or third fides, equal; and the two triangles fhall be equal; and their other angles fhall be equal, each to each, viz. those 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,

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For if the triangle ABC be applied to DEF fo 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, because the angle BAC is equal to the angle EDF/wherefore also the point C fhall coincide with the point F, because the straight line AC is equal to DF 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 does not coincide with the base EF, two straight lines would inclose a space, which is impoffible. There- a. 10. Ax. fore the base BC fhall coincide with the base EF, and be equal to it. Wherefore the whole triangle ABC fhall coincide with the whole triangle DEF, and be equal to it; and the other angles of the one 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 triangles have two fides of the one equal to two fides of the other, each to each, and have likewise the angles contained by those fides equal to one another; their bafes shall likewife be equal, and the triangles be equal, and their other angles to which the equal fides are oppofite, fhall be equal, each to each. Which was to be demonstrated.

PROP. V. THEOR.

THE angles at the bafe of an Ifofceles triangle are e

qual to one another; and if the equal fides be produced, the angles upon the other fide of the base shall be equal.

Let ABC be an Ifofceles triangle, of which the fide AB is equal

Book I. to AC, and let the straight 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.

a. 3. 1.

b. 4. 1.

In BD take any point F, and from AE, the greater, cut off AG equal to AF, the less, and join FC, GB.

A

Because 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 base FC is equal to the bafe 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 F angle AFC to the angle AGB. and because the whole AF is equal to the

b

whole AG, of which the parts AB, AC

B

C

G

E

C.

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c. 3. Ax. are equal; the remainder BF fhall 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 bafe 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 oppofite. therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. and fince it has been demonftrated that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are alfo 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 fide of the base. Therefore the angles at the base, &c. Q. E. D.

COROLLARY. Hence every equilateral triangle is alfo equiangular.

PROP. VI. THEOR.

IF two a be que oppofite to, the equal

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

angles fhall be equal to one another.

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