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UPON

Let ABC be a triangle having the angle ABC equal to the angle Book I. 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 o.
ther, let AB be the greater, and from it cut off DB equal to AC, the a. 2.1.
less, and join DC, therefore because in the

A
triangles DBC, ACB, DB is equal to AC,
and BC common to both, the two sides D
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 • ACB, the

b. 4. 1. B

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less to the greater ; which is abfurd. There-
fore 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.

PROP. VII. THEOR.
N 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 equai
to one another, and likewise those which are terminated
in the other extremity.

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

C D
Join CD; then, in the case 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 to the angle ADC. but the
angle ACD is greater than the angle
BCD, therefore the angle ADC is great. A

B
er 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 to the angle BCD; but it has
been demonstrated to be greater than it; which is impoflible.

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a. 5.1.

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d. 5. I.

Book I. But if one of the Vertices, as D, be within the other triangle
MACB; produce AC, AD to E, F. therefore
because AC is equal to AD in the triangle

E
ACD, the angles ECD, FDC upon the o-
ther side of the base CD are equal to
one another ; but the angle ECD is great-
er 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 e: A
qual to DB, the angle BDC is equal to
the angle BCD; but BDC has been proved to be greater than the
faine BCD, which is impossible. The case in which the Vertex of
onc 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 likewise those which are terminated in the other extremity. Q.E.D.

PROP. VIII. THEOR,
Ftwo triangles have two sides of the one equal to two

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bates equal; the angle which is contained by the two fides of the one shall be equal to the angle contained by the two lides equal to them, of the other.

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

D G
base BC equal to
the base EF. The
angle BAC is e-
qual to the angle
EDF.

For if the tri-
angle ABC be ap- B

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F plied to DEF so that the point B be on E, and the straight line BC upon EF; the point ( shall also coincide with the point F, because BC is equal to EF.

therefore BC coinciding with EF, BA and AC shall coincide with ED Book 1. and DF. for if the base BC coincides with the base EF, but the sides v 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 fides terminated in the other extremity. but this is impossible'. therefore if the base BC coincides with the base EF, the a. 7. 1. sides BA, AC cannot but coincide with the sides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal b to it. therefore if two triangles, &c. Q. E. D.

b. 8. Ax,

,

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PROP. IX. PROB.
O 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 bisect it.

Take any point D in AB, and from AC cut off AE equal to 2. 3. 1. AD; join DE, and upon it defcribeban

b. 1. 1.

А equilateral triangle DEF, then join AF. the straight line AF bisects the angle BAC. Because AD is equal to AE, and AF

D E is common to the two triangles DAF, EAF; the two sides DA, AF are eqnal to the two sides E A, AF, each to each; and B

F the base DF is equal to the base EF; therefore the angle DAF is equal to the angle EAF. wherefore the c. 8. 1. given rectilineal angle BAC is bifected by the straight line AF. Which was to be done.

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,

PROP. X. PROB.
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 b the a. 1.1

a angle ACB by the straight line CD. AB is cut into two equal parts b. 9. 1. in the poiot D.

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b. 1.1.

PROP. XI. PROB.
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 Din AC, and make CE equal to CD, and upon
DE describe b the equilateral tri-

F
anyle DFE, and join FC. the
straight line FC drawn from
the given point C, is at right
angles to the given straight line
AB.

Because DC is equal to CE,
and FC cominon to the two tri- A

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angles DCF, ECF; the two 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 e-

.qual to one another, each of them is called a right dangle; therefore d. 10. Def.

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.

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 PE at right angles to AB; and because ABC is a straight line, the angle

E B

c. $. I.

a. 10. Def.

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CBE is equal to the angle EBA;

Book 1. in the same manner, because ABD

E is a straight line, the angle DBE is equal to the angle EBA, wherefore the angle DBE is equal to the angle CBE, the less to the great

D er; which is impossible. therefore two straight lines cannot have a

A B common segment.

PROP. XII. PROB.
O draw a straight line perpendicular to a given

straight line of an unlimited length, from a given point without it.

Let AB be the given straight line, which may be produced to any length both ways, and let C be a point without it. It is required to draw a straight line perpendicu

C lar to AB from the point C.

Take any point D upon the other side of AB, and from the

E center C, at the distance CD, describe the circle EGF meet

H

b. 3. fot. ing AB in F, G; and bisect A • FG in H, and join CF, CH, CG, the straight line CH drawn from the given point C, is perpendicular to the given straight line AB.

Because FH is equal to HG, and HC common to the two triangles FHC, GHC, the two sides FH, HC are equal to the two GH, HC, each to each ; and the base CF is equal to the base CG; therefore d. 15. Det. the angle CHF is equal to the angle CHG; and they are adjacent angles. but when a straight line standing on a straight line makes the 8.8. adjacent angles equal to one another, each of them is a right angle, and the straight line which stands upon the other is called a perpendicular to it. therefore from the given point C a perpendicular CH has been drawn to the given straight line AB. Which was to be done.

PROP. XIII. THEOR.
THE angles which one straight line makes with ano-

ther upon one side of it, are either two right angles, or are together equal to two right angles.

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B

C. TO, I.

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