Let ABC be a triangle having the angle ABC equal to the angle Book I. ACB; the fide 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 equal to AC, the a. 1. less, and join DC. therefore 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

A

D

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

less to the greater; which is abfurd. There

Cb. 4. 1.

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.

U

PROP. VII. THEOR.

TPON the same base, and on the same side of it, See N. 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 upon the faine base AB, and upon the same side of it, which have their fides 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...

Join CD; then, in the cafe 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 er also than BCD; much more then is

CD

a. 5. г.

B

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.

Book I.

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

qual to DB, the angle BDC is equal to

the angle BCD; but BDC has been proved to be greater than the fame BCD, which is impossible. The cafe 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 fides 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,

F two triangles have two fides of the one equal to two fides of the other, each to each, and have likewise their. bafes 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.

DG

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 DF; and also the A base BC equal to the base EF. The angle BAC is e

plied to DEF fo

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 AC shall coincide with ED Book I.
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 same base EF and upon the fame
side 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
likewife their fides terminated in the other extremity. but this is im-
possible. therefore if the base BC coincides with the base EF, the a. 7. 1.
fides BA, AC cannot but coincide with the fides ED, DF; where-
fore likewise the angle BAC coincides with the angle EDF, and is

equal to it, therefore if two triangles, &c. Q. E. D.

PROP. IX. PROB.

T

O bisect

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

b. 8. Ax.

Let BAC be the given rectilineal arigle, it is required to bisect it.
Take any point D in AB, and from AC cut off AE equal to a. 3. 1.

AD; join DE, and upon it defcribeban

equilateral triangle DEF, then join AF.

A

b. 1. 1.

therefore the angle DAF is equal to the angle EAF. wherefore the c. 8. 1. given rectilineal angle BAC is bisected by the straight line AF.

Which was to be done.

T

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

1

Book I.

し

C. 4. 8.

See N.

2.3.1.

b. r. r.

c. 8. 1.

d. 10. Def.

1.

Because AC is equal to CB, and CD common to the two triangles ACD, BCD; the two fides AC, CD are equal to BC, CD, each to each; and the angle ACD is equal to the angle BCD; therefore the base AD is equal to the base DB, and the straight line AB is divided into two equal parts in the point D. Which was to be done.

T

A

PROP. XI. PROB.

O draw a straight line at right angles to a given straight line, from a given point in the fame. :

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.

F

Take any point Din AC, and make CE equal to CD, and upon
DE describe the equilateral tri-
angle 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 tri- AD

angles DCF, ECF; the two fides

C

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 is called a right dangle; 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.

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

If it be poffible, 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, the angle CBE is equal to the angle EBA; in the fame manner, because ABD 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 greater; which is impossible. therefore

two straight lines cannot have a common fegment.

T

A

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 perpendicular to AB from the point C.

C

Book I.

a. 10. Def. 1.

CG. the straight line CH drawn from the given point C, is per

pendicular to the given straight line AB.

e

1.

Because FH is equal to HG, and HC common to the two triangles FHC, GHC, the two fides FH, HC are equal to the two GH, HC, each to each; and the base CF is equal to the base CG; therefored. 15. Def. 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 C. 8. 1. 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 one fide of it, are either two right angles,

or are together equal to two right angles.