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Let ABC be a triangle having the angle ABC equal to the Book L 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 off DB. 3. 1. equal to AC, the less, and join DC; there

А fore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to

D both, the two sides 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 o ACB, the less to the great

b 4. 1. er: which is absurd. Therefore AB is not unequal to AC, that is, it is equal to it.

B

с Wherefore, if two angles, &c, Q. E. D.

Cox, Hence every equiangular triangle is also equilateral

PROP. VII. THEOR.

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 Soc Note. 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, sides CA, DA, terminated in the extremity A of the base equal to one another, and

C D likewise their sides CB, DB, that are terminated 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 equal to AD, the angle' ACD is equal a to the angle ADC: but the angle ACD is greater than the angle BCD; therefore the A А

B angle 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 fequal a to the angle BCD; but it has been demonstrated to be a 5. I greater than it ; which is impossible.

Book L 1

But if one of the vertices, as D, be within the other triangle.
ACB; produce AC, AD to E, F ; there-
fore, because AC is equal to AD in the

F
triangle ACD, the angles ECD, FDC

upon the other side of the base CD are a 5. 1.

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 than is
the angle BDC greater then the angle
BCD. Again, because CB is equal to
DB, the angle BDC is equal to the
angle BCD; but BDC has been prored A
to be greater than the same BCD ; which is impossible. The
case in wbich the vertex of one triangle is upon a side of the
other, needs no demonstration.

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.

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+ 1.3

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

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 ap.
plied to DEF, so B

CE

F
that the point B be on E, and the straight line BC upon
the point C shall also coincide with the point F. Because

EF;

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 impossible ; . therefore, if the basc BC coincides with a 7.1. the base ÉF, 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 b 8. Ax. triangles, &c. Q. E. D.

PROP. IX. PROB.

sect it.

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.

Take any point Din AB, and from AC cut off AE equal to a 3. 1. AD; join DE and upon it describe b

А

b1.1. an equilateral triangle DEF; then join AF; the straight line AF bisects the angle BAC. Because AD is equal to AE, and

D

E AF is common to the two triangles DAF, EAF; the two sides DA, AF, are equal to the sides EA, AF, each to each; ne base DF is equal to the

"'; therefore the angle DAF is ty. c to the angle B

Cc&L EAF; wherefore the given rectilineal angle BAC is bisected by the straight line AF, which was to be done.

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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 upon it an equilateral triangle ABC, and bisect 2 1.1. the angle ACB by the straight line CD. AB is cut into two b 9.1. equal parts in the point D.

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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: See Note. 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 23. I. upon DE describe the equi

F b1.1

lateral triangle 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 common to the two

sein, triangles DCF, ECF; the two

A D

E B sides DC, CF, are equal to the two EC, Okkie vach to each; and

the base DF is equal to the base EF; therefore the angle DCF 8. 1.

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 is

called a right d angle; therefore each of the angles DCF; ECF, d 10. Def. 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 poirt B draw BE at right angles to AB; and because ABC is a straight

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TO draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it.

A

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

с perpendicular to AB from the point C.

Take any point D upon the other side of AB, and from the centre C, at the distance

H CD, describe b the circle FDG

AF

b 3. Post.

GB meeting AB in F, G; and, bi

D sect FG in H, and join CF,

c 10.1. CH, CG; the straight line CH, drawn from the given point C, is perpendir to the given straight line AB.

Because qual to HG, and HC coinmon to the two trí. angles FHC, e two sides FH, HC are equal to the two GH, HC, each n; and the base CF is equal d to the base d 15, Def. CG; therefore the angle CHF is equale to the angle CHG; and 1. they are adjacent angles; but when a straight line standing on a e 8. 1. straight line makes the 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 another upon the one side of it, are either two right angles, or are together equal to two right angles.

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