g what is DCB; but the Angle ABF ACD. Thereproved before the Angle ABC ACB. Q. E. D.

fore.

COROLLART

Hence, every equilateral Triangle is alfo equiangular.

B

D

A

PROP. VI. ·

C

If two Angles (ABC, ACB) of a Triangle (ABC) be equal the one to the other; the Sides AB, AC, fubtended under the equal Angles, fhall also be equal.

the one to the other:

If the Sides are not equal, let either of them be the greater, as let BACA; now make "BD= CA, and draw the Line CD.

b

In the Triangles DBC, ACB, because BD= CA, and the Side BC is common; and the Angle DBCa ACB, the Triangles DBC and ABC fhall be equal to each other; and fo the Part is equal to the Whole: which is fabfurd.

с

CORO L.

Hence every equiangular Triangle is alfo equilateral.

PROP. VII.

If from the Extremes (A, B) of a right Line, two right Lines (AC, BC) be drawn to any Point C; you · cannot draw two other right Lines (AD, BD) on the fame fide the Point C from the fame Points A, B, to any other Point, as D; fo that AD, BD, be each equal to AC, BC, (that is, that AD be= AC, and BD BC) but only to C.

Cafe

a 9 ax.

Cafe 1. If the Point D falls in AC, it is manifeft that AD is not equal to AC. Cafe 2. If the Point D falls within the Triangle ACB, then draw the Line CD, and continue out BDF and BCE. Now if you fay that AD AC, then will the Angle ADC b beb 5. 1. ACD. Alfo fince BD BC, then fhall the c fup.

=

с

Angle FDC be ECD. Therefore the An-d 9 ax. gled FDCACD; that is, the Angle & FDC ADC: which is abfurd.

Cafe 3. If the Point D falls without the Triangle ACB, then join CD.

e

Again, the Angle ACD = ADC, and BCD e 5. 1. BDC; but the Angle ADC → BDC; that f 9 ax. js, the Angle ACDBCD; which is abfurd.

PROP. VIII.

If two Triangles (ABC, abc) have two Sides (AB, AC) each equal to two Sides, ab, ac, and the Bafe BC equal to the Bafe bc; then the Angles contained under the equal right Lines fhall be equal, viz. Aa.

AA

Because

b

Because a BC bc, if the Bafe BC be laid on the Base bc, they will coincide: therefore fince. AB ab, and AC ac, the Point A will fall on a, (for it cannot fall on any other Point, by the laft Propofition ;) and fo the Sides of the Angles A and a do coincided: wherefore those Angles are equal. Q. E. D.

COROLLARIES.

1. Hence Triangles mutually equilateral to one another, are also mutually f equiangular. 2. Triangles mutually equilateral, are equal the one to the other.

D

i I. I.

k conftr.

1 8. 1.

B

F

E

C

Take h AD AE,and draw the Line DE, upon which make an equilateral Triangle DFE: then draw the right Line AF, and it fhall bisect the Angle BAC.

For AD AE, and the Side AF is common, and the Bafe k DF-FE: therefore the Angle DAF EAF. Q. E. F.

Hence appears

COROL.

the way

of dividing an Angle into thefe equal Parts, viz. 4, 8, 16, &c. which is done by a new bifecting of each of the for

mer ones.

The Method of cutting Angles into any Number of equal Parts required by a Ruler and Compaffes, is as yet unknown to Geometri

cians.

PROP.

bifect the given Line AB.

For AC BC, and the Side CD is common, and the Angle ACDBCD: ThereforeAD BD. Q. E. F.

The Conftruction of Prop. 1. of this Book is fufficient for fhewing the Practice of this, and the laft Problem.

F

PROP. XI.

A

B

D

C

E

From a given Point

(C) in a given right
Line (AB), to erect a
right Line CF, at right

Angles to AB.

On either fide of the given Point take the Line CD = CE; upon the right Line DE erect fan equilateral Triangle: and draw the Line FC; and it will be the Perpendicular required.

a I. I.

e 3.1.

f I. I.

For CECD, and CF is common, and DF =EF (by Conft.) whence the Triangles DFC, EFC are mutually equilateral; therefore & the SS. 1. Angle DCF = ECF; whence FC is a Perpen-h 10 def. dicular. Q. E. F.

This and the following Problem are easily performed by means of a Square.

PROP.

XII.

To draw a right Line (CG) perpendicular upon (or to) a given infinite right Line (AB) from a given Point (C) without that Line.

C 2

With

c conftr.

which will be the Perpendicular required. Draw the Lines CE, CF; then the Triangles EGC, FGC are mutually equilateral: Therefore the Angles EGC, FGC are equal, and by e 10 def. confequence right ones: wherefore GC is a Perpendicular. QE.F

d 8. 1.

If the Angles ABC,

XIII.

When a right Line AB ftanding upon a right Line CD, makes with it Angles ABC, ABD; thofe are either two right Angles, or both together equal to two right Angles. ABD be equal; then it

f 10 def. is manifeft they make two right Angles; if unequal, then from the Point B let there be erectg 11.1. ed & a Perpendicular BE: because the Angle h 19 ax. ABC to a right Angle +ABE; therefore i 2 ax. ABC+ABD to one right Ang. + ABE k conft. +ABD to two right Angles. QE. D.

PROP. XIV.

A

If two right Lines (CB, BD) drawn contrary ways to the Point B in any ftrait Line

E (AB) Jo that the Angles (ABC, ABD) on each fide AB be equal to two right Angles: the faid Lines CB, BD, lie both in

C

B

D

the