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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, e. 15. Def. DL is equal to DG, and DA, DB, parts of them, are equal, 1.3. Ax. therefore the remainder AL is equal to the remainder f BG: But it has been fhown, 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 ftraight line BC. Which was to be done.

8. 2. I.

PROP. III. PRO B.

'ROM the greater of two given straight lines to cut off a part equal to the less.

FR

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.

From the point A draw a the ftraight line AD equal to C; and from the centre A, and at the dif

b. 3. Poft. tance AD, defcribe b the circle DEF; and because A is the centre

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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 straight line AE is equal to c C, and from AB, the greater of two straight lines, a part AE has been cut off equal to C the leis. Which was to be done.

I

PROP. IV. THEOREM.

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; they fhall likewife have their bafes, or third fides, 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

Book I.

D

ДА

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 triangle DEF; and
the other angles, to which
the equal fides are oppofite,
fhall be equal each to each,
viz. the angle ABC to the B
angle DEF, and the angle
ACB to DFE.

CE

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 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 the straight line AC is equal to DF: But the point B coincides with the point E; wherefore the bafe BC fhall coincide with the bafe EF; because the point B coinciding with E, and C with F, if the base BC does not coincide with the bafe EF, two ftraight lines would inclofe a fpace, which is impoffible a. Therefore a 10 Ax. the base BC fhall coincide with the bafe 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 fhall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to 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 thofe fides equal to one another, their bases shall likewise 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 demonftrated.

THE

PROP. V. THEOR.

THE angles at the base of an Ifofceles triangle are equal to one another; and, if the equal fides be produced, the angles upon the other fide of the base thall be equal.

Let ABC be an Ifofceles triangle, of which the fide AB is e-
B

qual

Book I, qual 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.

In BD take any point F, and from AE the greater, cut off

a 3. 1. AG equal a to AF, the lefs, and join FC, GB.

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 com

F

B

A

G

E

mon to the two triangles AFC, AGB 3; therefore the base FC is eb 4. 1 qual b to the bafe GB, and the triangle 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 the equal fides are oppofite; 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; the c 3. Az. 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 b, 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 demonstrated, 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 bafe of the triangle ABC: And it has also been proved that the anlge FBC is equal to the angle GCB, which are the angles upon the other fide of the bafe. Therefore the angles at the base, &c. Q. E. D.

COROLLARY. Hence every equilateral triangle is also equiangular. PROP. VI. THE OR.

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

Let

19.

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

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 a off DB e a 3. I. qual to AC, the less, and join DC; there

fore, 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; and the
angle DBC is equal to the angle ACB;
therefore the base DC is equal to the base
AB, and the triangle DBC is equal to
the triangle b ACB, the less to the great-
B
er; which is abfurd. Therefore AB is
not unequal to AC, that is, it is equal to

it. Wherefore, if two angles, &c. Q. E. D.

A

D

COR. Hence every equiangular triangle is also equilateral.

UPON

PROP. VII. THEOR.

b 4. I.

PON the fame base, and on the fame fide of it, See N. there cannot be two triangles that have their fides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity.

If it be poffible, let there be two triangles ACB, ADB, upon the fame base AB, and upon the fame fide of it, which have their fides CA, DA, terminated in the extremity A of the base equal to one another, and likewise their fides CB, DB, that are terminated in B.

Join CD; then, in the cafe in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle, ACD is equal a to the angle

C

D

ADC: But the angle ACD is greater

than the angle BCD; therefore the A angle ADC is greater also than BCD;

a s. I.

B

a to

much more then is the angle BDC greater than the angle BCD.
Again, because CB is equal to DB, the angle BDC is equal
the angle BCD; but it has been demonftrated to be greater
than it; which is impoffible.

B 2

But

Book. I.

E

F

But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F; therefore because AC is equal to AD in the triangle ACD, the angles ECD, FDC upon the other fide of the base CD are a 5. i. equal a to one another, but the angle ECD is greater, than the angle BCD; wherefore the angle FDC is likewife greater than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to A DB, the angle BDC is equal a to the

angle BCD; but BDC has been proved to be greater than the fame BCD; which is impoffible. The cafe in which the ver tex of one triangle is upon a fide of the other, needs no demonftration.

Therefore, upon the fame bafe, and on the fame fide of it, there cannot be two triangles that have their fides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. Q. E. D.

I

PROP. VIII. THEOR.

F two triangles have two fides of the one equal to two fices of the other, each to each, and have likewise their bafes equal; the angle which is contained by the two fides of the one shall be equal to the angle contained by the two fides equal to them, of the other.

Let ABC, DEF be two triangles having the two fides AB, AC, equal to the two fides DE, DF, each to each, viz. AB te DE, and AC to A

DF; and also the
bafe BC equal to
the bafe EF. The
angle BAC is e-
qual to the angle
EDF.

For, if the tri-
angle, ABC be ap-

plied to DEF, fo

B

D G

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that the point B be on E, and the ftraight line BC upon EF; the point Chall also coincide with the point F. Because

BC

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