Book I.

e. 15. Def.

f. 3. Ax.

Because the point B is the center of the circle CGH, BC is equal to BG. and because D is the center 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 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 fame 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 straight line BC. Which was to be done.

ROM the greater of two given straight lines to eut off a part equal to the lefs.

FROM

b. 3. Poft.

from the center A, and at the diftance AD defcribe the circle DEF. and because A is the center of the circle DEF, AE shall be equal to AD. but the straight line C is likewise equal to AD. whence AE and C are each of c. 1. Ax. 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 less. Which was to be done.

F

I

F 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 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 each, viz. those to which the equal fides are opposite.

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,

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

For if the triangle ABC be applied to 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 ABis equal to DE. and AB coinciding with DE, AC shall coincide with DF, because the angle BAC full on is equal to the angle, EDF. wherefore also the point C shall coincide with the point F, because the ftraight 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 impossible. Therefore the base BC shall coincide with the a. 1o. Ax. base EF, and be equal to it. Wherefore the whole triangle ABC shall 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 ABCto 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 likewife the angles contained by those fides equal to one another; their bases shall likewife be equal, and the triangles be equal, and their other angles to which the equal fides are oppofite, shall be equal, each to each. Which was to be demonftrated.

T

PROP. V. THEOR.

A

THE 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 fide of the base shall be equal.

Let ABC be an Isosceles triangle, of which the side AB is equal

4

Book I. 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.

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.

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 tri

angles AFC, AGB; therefore the base
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 opposite; viz. the
angle ACF to the angle ABG, and the F
angle AFC to the angle AGB. and be-

B

A

C

G

cause the whole AF is equal to the D
whole AG, of which the parts AB, AC

E

c. 3. Ax. are equal; the remainder 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 FBCis 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 also 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 also equiangular.

I

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

:

Let ABC be a triangle having the angle ABC equal to the Book I. angle ACB; the side 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 off DB equal to AC, a. 3. 1. the less, and join DC. therefore because in

the triangles DBC, ACB, DB is equal to

A

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

D

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 greater; which B

C

b. 4. I.

is abfurd. Therefore AB is not unequal to AC, that is, it is
equal to it Wherefore if two angles, &c. Q. E. D.

Cor. Hence every equiangular triangle is also equilateral.

U

PROP. VII. THEOR.

PON the fame base, and on the same side 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 likewife 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 same side 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.

C

:

D

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 ADC. but the

a. 5. Ι.

angle ACD is greater than the angle

BCD, therefore the angle ADC is great

er also than BCD; much more then is

B

the angle BDC greater than the angle BCD. again, because ÇB

is equal to DB, the angle BDC is equal to the angle BCD; but it

has been demonftrated to be greater than it; which is impossible.

Book I.

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

2. 5. 1.

ACD, the angles ECD, FDC upon the
other fide of the base CD are equal to
one another; but the angle ECD is greater
than the angle BCD, wherefore the angle

CA

D

FDC is likewife greater than BCD; much
more then is the angle BDC greater than
the angle BCD. again, because CB is e-
qual to 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 impossible. The cafe in which the Vertex
of one triangle is upon a fide of the other, needs no demonstration.
Therefore upon the fame base, and on the same side 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 likewife

B

those which are terminated in the other extremity.

E. D.

PROP. VIII.

THEOR.

IF

F two triangles have two fides of the one equal to two fides of the other, each to each, and have likewife their bases 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.

DG

Let ABC, DEF be two triangles having the two fides AB, AC equal to the two fides DE, DF, each to each, viz. AB to DE, and AC to DF; and alfo the base BC equal A to the base EF. The angle BAC is equal to the angle EDF.

For if the trian

gle ABC be applied

to DEF so that the B

CE

F

point B be on E, and the straight line BC upon EF; the point C

shall alfo coincide with the point F, because BC is equal to EF.