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a 3. I.

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 BCE.

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 sides FA, AC are equal to the two GA, AB, each to each ; and they contain the angle FAG common to the two triangles AFC, A

AGB; therefore the base FC is eb. 4. I.

qual to the base GB, and the tri-
angle AFC to the triangle AGB;
and the remaining angles of the one
are equal o to the remaining angles
of the other, each to each, to which B
the equal fides are opposite; viz.
the angle ACF to the angle ABG,
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

E parts AB, AC are equal; the reC 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 bafe 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 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 demonstrated, that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF, are allo 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.

PRO P. VI. THEOR.
F 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.

Le

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Let ABC be a triangle having the angle ABC equal to the Book I. angle ACB; the side , B is also equal to the side AĆ.

For, if AB bę not equal to AC, one of them is greater than ibe other : Let AB be the greater, and from it cut off DB e- a 3. ti qual to AC, the less, and join DC; there

A fore, because in the triangles DBC, ACB, DB is equal to AC, and EC common to D 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 bale DC is equal to the base AB, and the triangle DBC is equal to the triangle b ACB, the less to the

b 4. 1 greater; which is absurd. Therefore B 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.

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PRO P. VII. THEO R.
ON the same base, and on the same side 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 likewise those which are terminated
in the other extremity.

If it be possible, let there be two triangles ACB, ADB, up-
on the same base AB, and upon the same side of it, which have
their lides CA, DA, terminated in the extremity A of the
bafe; equal to one another, and like:

С D
wise their fides CB, DB that are ter-
minated in B.

Join CD; then, in the case in
which the vertex of each of the tri-
angles is without the other triangle,
because AC is equalto AD, the angle
ACD is equal a to the angle ADC:

á š. I.
But the angle ACD is greater than
the angle BCD; therefore the angle

B
ADC is greater also 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 it has been demonitrated to be greater
than it; which is impoflible.

But

B 2

Book I.

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But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F;

E
therefore, because AC is equal to AD

F
in the triangle ACD, the angles ECD,
FDC upon the other side 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-
wise greater than BCD; much more
then is the angle BDC greater than the
angle
BCD. Again, because CB is equal A

B
to DB, the angle BDC is equal a to the
angle BCD; but BDC has been proved to be greater than the
same BCD, which is impossible. The case in which the ver-
tex of one triangle is upon a side of the other needs no de-
monstration.

Therefore upon the same base, and on the same fide 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. THE OR.

IF
F 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 con-
tained 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 co
DE, and AC to A

D G
DF; and also the
base BC equal to
the base EF. The
angle BAC is e.
qual to the angle
EDF.

For, if the tri.
angle ABC be B

CE

F
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 I. AC shall coincide with ED and DF; for, if the base BC coincides with the base EF, but the Gides 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 same 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 impoffible; therefore if the base BC coin. a 7.1. cides with the 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 b 8. Ax. if two triangles, &c.

Q. E. D.

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po 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 2-8 3. . qual to ĄD; join DE, and upon iç describe an equilateral triangle DEF;

А

bi. I then join AF; the straight line AF bifects the angle BAC.

Because AĎ is equal to AE, and AF is common to the two triangles D E DAF, EAF; the two sides DA, AF, are equal to the two fides EA, AF, each to each; and the base DF is e

BI qual to the base EF; therefore the F

с angle DAF is equal to the angle EXF; wherefore the given rectilineal angle BAC is bifected by the straight line AF. Which was to be done.

C 8. I.

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Ti

10 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 * s. !. the angle ACB by the straight line CD. AB is cut into two o g. 1, equal parts in the point D.

B 3

Because

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TO
o draw a straight line at right angles to a given

straight line, from a given point in the fame, See N.

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-

F
teral triangle DFE, and join
FC; the straight line FC drawn
from the given point C is at
right angles

the given
straight line AB.

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

two EC, CF, each to each ; and the base DT is equal to the C 8. 1.

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 ;

therefore each of the angles DCF, ICF 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.

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 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,

E B

1.

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