Book I. to the triangle b ACB, the lefs to the greater; which is abfurd. Therefore AB is not unequal to AC, that is, it is equal to it. b. 4. 1. Wherefore, if two angles, &c. Q. E. D.

See N.

I

COR. Hence every equiangular triangle is alfo equilateral.

I'

PROP. VII. THEOR.

F there be two triangles upon the fame base, and on the fame fide of it, that have their fides, which are terminated in one extremity of the base equal to one another, they shall not have their fides equal, which are terminated in the other extremity.

Let there be two triangles ACB, ADB, upon the fame bafe AB, and upon the fame fide of it, which have their fides CA, DA, terminated in the extremity A of the base, equal to one another; their fides CB, DB, that are terminated in B, fhall not be equal.

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 a to the angle ADC: But the
angle ACD is greater than the angle
BCD; therefore the angle ADC is
greater alfo than BCD; much more
then is the angle BDC greater than the
angle BCD. But if CB were equal to Α
DB, the angle BDC, would be equal a

C D

B

to the angle BCD; and it has been demonftrated to be greater than it; which is impoffible; therefore CB cannot be equal to DB.

"

F

But, if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F; therefore, because AC is equal to AD, in the triangle ACD, the angles ECD, FDC upon the other fide 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 likewife greater than BCD; much more then is the angle BDC greater than the angle BCD. But if CB were equal to DB, the angle BDC would be equal a to the angle. BCD; and BDC has been proved to be greater than the fame

A

B

BCD;

BCD; which is impoffible. Therefore CB cannot be equal to Book I. DB. The cafe in which the vertex of one triangle is upon a fide of the other, needs no demonftration. Wherefore, if there be two triangles, &c. Q. E. D.

PROP. VIII.

F two triangles have two

IF

THEOR.

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, fhall be equal to the angle contained by the two fides equal to them of the other.

A

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 alfo the bafe BC equal to the bafe EF. The angle BAC is equal to the angle EDF.

For, if the triangle ABC be applied to DEF, fo that the point B be on E, and the ftraight line BC upon EF; the point C fhall alfo coincide with the point F, B

because BC is equal to EF;

СЕ

G

F

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

PROP. IX. PROB.

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

Tit

Let BAC be the given rectilineal angle, it is required to bifect it.

BOOK I.

a

Take any point D in AB, and from AC equal to AD; join DE, and upon it de23. 1. ferine b an equilateral triangle DEF; then join b. 1. AF; the ftraight line AF bifects the angle BAC.

cut a off AE A

D

E

Because AD is equal to AE, and AF is common to the two triangles DAF, EAF; the two fides DA, AF are equal to the two fides EA, AF, each to each; and the bafe DF is alfo B

C

cs. 1. equal to the bafe EF; therefore the angle DAF is equal © to the angle EAF; wherefore the given rectilineal angle BAC is bifected by the ftraight line AF. Which was to be done.

a I. I.

b 9. I.

€ 4. I.

23.1.

b 1. I.

€ 8.1.

T

O bifect a given finite ftraight line, that is, to divide it into two equal parts.

Let AB be the given ftraight line; it is required to divide it into two equal parts.

Defcribe a upon it an equilateral triangle ABC, and bifect b the angle ACB by the ftraight line CD. AB is cut into two equal parts in the point D.

A

Becaufe 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
alfo equal to the angle BCD; therefore
the bafe AD is equal to the base c DB,
and the ftraight line AB is divided into
two equal parts in the point D. Which A
was to be done.

PROP. XI. PROB.

D

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

Let AB be a given ftraight line, and C a point given in it; it is required to draw a ftraight 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 defcribe b the equilateral triangle DFE, and join FC; the traight line FC drawn from the given point C, is at right angles to the given ftraight line AB.

Becaufe DC is equal to CE, and FC common to the two triangles DCF, ECF; the two fides DC, CF are equal to the two EC, CF, cach to each; and the base DF is also equal to the base EF; therefore the angle DCF is equal to the angle

C

ECF;

F

ECF; and they are adjacent angles. But,
when the adjacent angles which one
ftraight 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 is a right
angle. Wherefore, from the given point
C, in the given ftraight line AB, FC has been drawn at right
angles to AB. Which was to be done.

T&

PROP. XII. PROB.

A D

C

EB

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

C

It

Let AB be the given ftraight line, which may be produced to any length both ways, and let C be a point without it. is required to draw a ftraight line perpendicular to AB from the point C. Take any point D upon the other fide of AB, and from the centre C, at the distance CD, defcribe a the circle EGF meeting AB in F, G; and bifect b FG in H, and join CF, CH, CG; the ftraight line CH, drawn from the given point C, is perpendicular to the given ftraight line AB.

A F

E

a 3. Poft.

H

DG B

b 10. I.

Becaufe FH is equal to HG, and HC common to the two triangles FHC, GHC, the two fides FH, HC are equal to the

C

two GH, HC, each to each; and the base CF is alfo equal to c 15. Def. the base CG; therefore the angle CHF is equal d to the angled 8. 1. CHG; and they are adjacent angles; but when a straight line standing on a 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 ftraight line AB. Which was

to be done.

THE

PROP. XIII. THEOR.

HE angles which one ftraight line makes with another upon the one fide of it, are either two right angles, or are together equal to two right angles.

Let the ftraight line AB make with CD, upon one' fide of it,

the

BOOK I. the angles CBA, ABD; these are either two right angles, or are together equal to two right angles.

a Def. 10.

For, if the angle CBA be equal to ABD, each of them is a

right a angle; but, if not, from the point B draw BE at right b11. 1. angles b to CD; therefore the angles CBE, EBD are two right

angles a; and becaufe CBE is equal to the two angles CBA, ABE together, add the angle EBD to each of these equals; therefore c 2. Ax. the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. Again, because the angle DBA, is equal to the two angles DBE, EBA, add to thefe equals the angle ABC; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC; but the angles CBE, EBD have been demonftrated to be equal to the fame three angles; and things 1. Ax. that are equal to the fame, are equal d to one another; therefore the angles CBE, EBD are equal to the angles DBA, ABC: but CBE, EBD are two right angles; therefore, DBA, ABC are together equal to two right angles. Wherefore, when a straight line, &c. Q. E. D.

d 1.

a 13. I.

PROP. XIV. THEOR.

TF, at a point in a ftraight line, two other ftraight lines, upon the oppofite fides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the fame ftraight line.

At the point B in the straight line AB, let the two ftraight lines BC, BD, upon the oppofite fides of AB, make the adjacent angles ABC, ABD equal together to two right angles. BD is in the fame ftraight line with CB.

For, if BD be not in the fame C ftraight line with CB, let BE be in

B

A

E

D

the fame fraight line with it; therefore, becaufe the straight line AB makes angles with the ftraight line CBE, upon one fide of it, the angles ABC, ABE are together equal a to two right angles; but the angles ABC, ABD are likewise together

equal