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 BСЕ.

a 3. 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 com

b 4. I.

mon to the two triangles AFC,
AGB; therefore the base FC is e-
qual to the base GB, and the tri-
angle 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,

A

and the angle AFC to the angle F

G

AGB: And because the whole AF is
equal to the whole AG, of which the D
parts AB, AC are equal; the re-

E

• 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, since 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 bafe, &c. Q. E. D.

COROLLARY. Hence every equilateral triangle is also equiangular.

PROP. VI. THEOR.

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

Le

For, if AB be not equal

1

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.

to AC, one of them is greater than

i

the other: Let AB be the greater, and from it cut off DB e-
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 triangleb ACB, the less to the
greater, which is abfurd. Therefore B

a 3. 1.

A

D

A

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

PON the fame base, and on the fame fide of it, see Ni 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 fame fide of it, which have their fides CA, DA, terminated in the extremity A of the

base, equal to one another, and like

wise their fides CB, DB that are ter-
ininated in B.

C

D

ADC is greater also than BCD;

af. r.

B

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

B2

But

Book I.

~ angle ACB; produce AC, AD to E, F,

85. 1.

therefore, because AC is equal to AD
in the triangle 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 FDC is like-
wife greater than BCD; much more
then is the angle BDC greater than the
angle BCD. Again, because CB is equal A

But if one of the vertices, as D, be within the other tri

to DB, the angle BDC is equal to the

angle BCD; but BDC has been proved to be greater than the same BCD, which is impossible. The case 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 sides which are terminated in one extremity of the base equal to one another, and likewife those which are terminated in the other extremity, Q.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.

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 A

DF; and also the

base BC equal to

DG

the base EF. The

applied to DEF,

so that the point B be on E, and the straight line BC upon EF; the point C shall also coincide with the point F, because

BC

BC is equal to EF; therefore BC coinciding with EF, BA and Book J. AC shall coincide with ED and DF; for, if the base BC coincides with the base EF, but the fides BA, CA do not coincide with the fides ED, FD, but have a different situation, as EG, FG; then, upon the fame base EF, and upon the fame fide of it, there can be two triangles that have their fides which are terminated in one extremity of the base equal to one another, and likewise their fides terminated in the other extremity: But this is impossible; therefore. if the base BC coin- a 7.1. cides with the base EF, the fides BA, AC cannot but coincide with the fides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal to it. Therefore b 8. Ax. if two triangles, &c. Q. E. D.

T

PROP. IX. PRO B.

O bifect a given rectilineal angle, that is, to divide
it into two equal angles.

Let BAC be the given rectilineal angle, it is required to bi

fect it.

Take any point D in AB, and from AC cut off AE c-a3..

qual to AD; join DE, and upon iç

b

describe an equilateral triangle DEF;

A

then join AF; the straight line AF

bisects the angle BAC.

Because AD AD is equal to AE, and
AF is common to the two triangles D

E

DAF, EAF; the two fides DA, AF,

angle DAF is equal to the angle

EAF; wherefore the given rectilineal angle BAC is bisected

by the straight line AF. Which was to be done.

PROP. X. PROBВ.

O bisect a given finite straight line, that is, to divide
it into two equal parts.

Let AB be the given straight line; it is required to divide it

into two equal parts.

Describe upon it an equilateral triangle ABC, and bisect 1.1. the angle ACB by the straight line CD. AB is cut into two 9.1. equal parts in the point D.

B3

Because

See N.

a 3. 3.

b. 4.

O draw

linh
a ftraight line at right angles to a given

a given point in the same.

Let AB be a given straight line, and C a point given in it; it is required to draw a straight line from the point C at right angles to AB.

Take any point D in AC, and make CE equal to CD, and

upon DE describe the equila

teral triangle DFE, and join

FC; the straight line FC drawn

from the given point C is at

F

right angles to the given

straight line AB.

Because DC is equal to CE,
and FC common to the two
triangles DCF, ECF; the two A D

C

EB

c 8. 1.

fides DC, CF are equal to the two EC, CF, each to each; and the base DF is equal to the bafe EF; therefore the angle DCF is equal to the angle ECF; and they are adjacent angles. But, when the adjacent angles which one straight line makes with another straight line are d 10. Def. equal to one another, each of them is called a right d angle;

I.

therefore each of the angles DCF, ECF is a right angle. Wherefore from the given point C, in the given ftraight line AB, FC has been drawn at right angles to AB. Which was

to be done.

COR. By help of this problem, it may be demonftrated, that two straight lines cannot have a common fegment.

If it be possible, let the two straight lines ABC, ABD have the fegment AB common to both of them. From the point B draw BE at right angles to AB; and because ABC is a straight line,