Book I. equal 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 equala to AF, the less, and join FC, GB.

Because AF is equal to AG, and AB to AC, the two sides FA, AC are equal to the two GA, AB, each to each; and they contain the angle FAG com

mon to the two triangles AFC,
AGB; therefore the base FC is
equal 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 sides are opposite; 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

F

D

A

B

C

G

E

c 3. Ax. remainder BF shall be equal to the remainder CG; and FC was proved to be equal to GB; therefore the two sides 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 equalb, and their remaining angles, each to each, to which the equal sides 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 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 side of the base. Therefore the angles at the base, &c. Q. E. D. COROLLARY. Hence every equilateral triangle is also equiangular.

PROP. VI. THEOR.

IF two angles of a triangle be equal to one another, the sides also which subtend, or are opposite 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 side 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 a 3. 1.

equal to AC, the less, and join DC; there

A

D

fore, because in the triangles DBC, ACB,
DB is equal to AC, and BC common to
both, the two sides DB, BC are equal to
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 ACB, the less to the great-
er; which is absurd. Therefore AB is
not unequal to AC, that is, it is equal to
it. Wherefore, if two angles, &c. Q. E. D. B
Cor. Hence every equiangular triangle is also equilateral.

PROP. VII. THEOR.

b 4. 1.

C

UPON the same base, and on the same side of it, See N. there cannot be two triangles that have their sides 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 possible, let there be two triangles ACB, ADB, upon the same base AB, and upon the same side of it, which have their sides CA, DA terminated in the extremity A of the base equal to one another, and likewise their sides CB, DB that are terminated in B.

CD

angle ADC is greater also than BCD;

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 a to a 5. 1. the angle BCD; but it has been demonstrated to be greater than

it, which is impossible.

Book I.

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

a 5. 1.

ACB; produce AC, AD to E, F; there-
fore, because AC is equal to AD in the
triangle ACD, the angles ECD, FDC
upon the other side of the base CD are
equal a to one another, but the angle
ECD is greater than the angle BCD;
wherefore the angle FDC is likewise
greater than BCD; much more then is
the angle BDC greater than the angle
BCD. Again, because CB is equal to
DB, the angle BDC is equal a to the angle
BCD; but BDC has been proved to be A
greater than the same BCD; which is impossible. The case in
which the vertex of one triangle is upon a side of the other, needs
no demonstration.

E

F

C

D

B

Therefore upon the same 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 likewise those which are terminated in the other extremity. Q. E. D.

PROP. VIII. THEOR.

IF two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them, of the other.

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

DF; and also the

base BC equal to

DG

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 is equal to EF; therefore BC coinciding with EF, BA and Book I. AC shall coincide with ED and DF; for, if the base BC coincides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but have a different situation, as EG, FG; then, upon the same base EF, and upon the same side of it, there can be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity; but this is impossiblea; therefore, if the base BC coincides with the a 7. 1. base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore, likewise, the angle BAC coincides with the angle EDF, and is equal to it. Therefore, if two triangles, b 8. Ax. &c. Q. E. D.

PROP. IX. PROB.

TO bisect 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

sect it.

Take any point D in AB, and from AC cuta off AE equal to a 3. 1.

AD; join DE, and upon it describe b an equilateral triangle DEF; then join AF; the straight line AF bisects the angle BAC.

A

b 1. 1.

Because AD is equal to AE, and AF is common to the two triangles DAF, EAF; the two sides DA, AF are equal to the two sides EA, AF, each to each; and the base DF is

D

E

equal to the base EF; therefore the

F

angle DAF is equal to the angle B

Cc8.1

EAF; wherefore the given rectilineal angle BAC is bisected by

the straight line AF. Which was to be done.

PROP. X. PROB.

TO 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 a upon it an equilateral triangle ABC, and bisect ba 1. 1. the angle ACB by the straight line CD. AB is cut into two b 9. 1. equal parts in the point D.

See N.

a 3. 1.

b 1. 1.

PROP. XI. PROB.

TO draw a straight line at right angles to a given straight line, from 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 a make CE equal to CD, and upon DE describe b the equi

lateral triangle DFE, and join

F

FC; the straight line FC drawn

from the given point C is at

right angles to the given straight
line AB.

c 8. 1.

1.

Because DC is equal to CE, and FC common to the two triangles DCF, ECF; the two A D sides DC, CF are equal to the two EC, CF, each to each; and the base DF is equal to the base 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 equal to one another, each of them 10. Def. is called a right d angle; therefore each of the angles DCF, ECF is a right angle. Wherefore, from the given point C, in the given straight line AB, FC has been drawn at right angles to AB. Which was to be done.

C

EB

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

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