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late of Euclid allows the doing of this, as Proclus well observes.

PROP. III.

Two right Lines (A and BC) being given; from the greater (BC) to take away a right Line (BE), equal to the lesser (A.)

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a 2. Ι.

b 15 def.

c conft. d 1 ax.

From the Point B draw the right Line BD A; then a Circle described about the Center B, with the Distance BD, shall cut off BE=BD=A=BE. Which was to be

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If two Triangles (BAC, bac) have two Sides of the one (BA, AC) equal to two Sides of the other (ba, ac) each to its correspondent Side (that is, BA =ba, and AC=ac) and have one Angle (A) equal to one Angle (a), contained under the equal right Lines: they shall have the Base (BC) equal to the Base (bc), and the Triangle (BAC) shall be e

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qual to the Triangle (bac); and the remaining Angles (B, C) shall be equal to the remaining Angles (b,c) each to each, under which the equal Sides are Subtended.

If the Point a be applied to the Point A, and the right Line ab placed upon the right Line AB, the Point b shall fall upon B, because ab = AB: alfo the right Line ac shall fall upon AC, because the Angle A = a; moreover, a kyp. the Point c shall fall upon the Point C, because AC ac. Therefore the right Lines bc, BC, because they have the fame Bounds, shall agree bor coincide, and so confequently are equal. b 8 ax. Wherefore the Triangles BAC, bac, and the Angles B, b, as also the Angles C, c, do agree, or coincide, and are equal. Which was to be demonStrated.

PROP. V.

A

The Angles (ABC, ACB) at the Base of an Isofceles Triangle are equal to each other. And if the equal right Lines AB, AC, be continued out, the Angles (CBD, BCE) under the Base, are equal to

D

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E

each other.

2

Take AE = AD, and a 3. 1. join C, D, and B, E.b 1 poft. Then in the Triangles c byp. ACD, ABE, because AB is = AC, and A Edd conft. AD, and the Angle A is common : therefore shall the Angle ABE be = ACD, and the e 4. 1. Angle AEB = ADC, and the Bafe BE=DC. Also EC isf DB. Whence in the Triangles f 3 ax. BCE, BDC, the Angle ECB shall be = DBC, (which is the latter part of the Propofition to be demonstrated) and confequently the Angle EBC = DCB;

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g what is DCB; but the Angle ABF = ACD. Thereproved be- fore the Angle ABC = ACB. Q. E. D.

fore.

h 3 ax.

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COROLLARY.

Hence, every equilateral Triangle is alfo equiangular.

D

A

PROP.

A

B

C

VI.

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, shall 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.

In the Triangles DBC, ACB, because BD= CA, and the Side BC is common; and the Angle DBC = ACB, the Triangles DBC and ABC shall be equal to each other; and so thể Part is equal to the Whole: which is fabsurd.

COROL.

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 Same fide the Point C from the Same Points A, B, to any other Point, as D; So that AD, BD, be each equal to AC, BC, (that is, that AD be = AC, and BD = BC) but only to C.

Cafe

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a 9 ax.

the Point D falls in AC, it is t AD is not equal to AC. the Point D falls within the Trithen draw the Line CD, and conOF and BCE. Now if you say that then will the Angle ADCbbe = b 5.1. fince BD = BC, then shall the c fup. be=ECD. Therefore the An-d 9ax. -ACD; that is, the Angle FDC which is abfurd.

the Point D falls without the Trithen join CD.

me Angle ACD = ADC, and BCDe 5.1. but the Angle ADC → BDC; that f 9 ax. le ACD BCD; which is abfurd.

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1

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AB = ab, and AC = ac, the Point A will fall on a, (for it cannot fall on any other Point, by the last Propofition ;) and so 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

B

A

F

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For AD=AE, and the Side AF is common, and the Base * DF = FE : therefore the Angle DAF = EAF. Q. E. F.

COROL.

Hence appears the way of dividing an Angle into these equal Parts, viz. 4, 8, 16, &c. which is done by a new bisecting 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

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cians.

To drar (or to

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