XXIV. Of three-sided Figures that is an Equilateral Triangle, which hath three equal Sides. XXV. That an Isofceles, or Equicrural one, which bath only two Sides equal.

XXVI. And a Scalene one, is that, which hath three unequal Sides.

XXVII. Also of three-sided Figures, that is a Right-angled Triangle, which bath a Right Angle.

XXVIII. That an Obtuse-angled one, which hath an Obtuse Angle.

XXIX. And that an Acute-angled one, which hath three Acute Angles.

XXX. Of four-fided Figures, that is a Square, whose four Sides are equal, and its Angles all Right ones.

XXXI. That an Oblong, or Rectangle, which is longer than broad; buti its opposite Sides are equal, and all its Angles Right ones. XXXII. That a Rhombus, which bath four equal Sides, but not Right Angles. XXXIII. That a Rhomboides, whose oppofite Sides and Angles only are equal. XXXIV. All Quadrilateral Figures, besides these, are called Trapezia.

XXXV. Parallels are fuch Right Lines, in the Same Plane, which, if infinitely produc'd buth Ways, would never meet.

POSTULATES.

1. GRANT that a Right Line may be drawn from any one Point to another.

II. That a finite Right Line may be continued directly forwards.

III. And that a Circle may be described about any

Centre with any Distance.

B2

AXIOM Ş.

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ΑΧΙΟMS.

THINGS equal to one and the same Thing, are equal to one another.

II. If to equal Things are added equal Things, the Wholes will be equal.

III. If from equal Things equal Things be taken away, the Remainders will be equal.

IV. If equal Things be added to unequal Things, the Wholes will be unequal.

V. If equal Things be taken from unequal Things,
• the Remainders will be unequal.

VI. Things which are double to one and the same
Thing, are equal between themselves.
VII. Things which are half one and the fame
Thing, are equal between themselves.
VIII. Things which mutually agree together, are
equal to one another.

IX. The Whole is greater than its Part.
X. Two Right Lines do not contain a Space.
XI. All Right Angles are equal between them-
selves.

XII. If a Right Line, falling upon two other
Right Lines, makes the inward Angles on the
fame Side thereof, both together, less than two
Right Angles, those two Right Lines, infinitely
produc'd, will meet each other on that Side
where the Angles are less than Right ones.

Note, When there are several Angles at one Point, any one of them is express'd by three Letters, of which that at the Vertex of the Angle is plac'd in the Middle. For Example; in the Figure of Prop. XIII. Lib. I. the Angle contain'd under the Right Lines AB, BC, is called the Angle ABC; and the Angle contained under the Right Lines A B, BE, is called the Angle A BE.

PRO

PROPOSITION I.

PROBLEM.

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To defcribe an Equilateral Triangle upon a given finite Right Line.

L

ET AB be the given finite Right Line, upon
which it is required to describe an equilateral
Triangle.

About the Centre A, with the Distance AB, describe the Circle BCD*; and about the Centre B,* Poft. 3 with the fame Distance BA, describe the Circle ACE*; and from the Point C, where the two Circles cut each other, draw the Right Lines CA, CB+.

† Poft. 1.

Then because A is the Centre of the Circle D BC, A C shall be equal to A B‡. And because B is the Def. 15. Centre of the Circle CAE, B C shall be equal to BA: But CA hath been proved to be equal to AB; therefore both CA and CB are each equal to A B. But Things equal to one and the fame Thing, are equal between themselves *, and confequently A C is * Ax. 1. equal to CB; therefore the three Sides CA, А В, BC, are equal between themselves.

And so, the Triangle BAC is an Equilateral one, and is described upon the given finite Right Line AB; which was to be done.

PROPOSITION II.

PROBLEM.

At a given Point to put a Right Line equal to a
Right Line given.

LET the Point given be A, and the given Right
Line B C; it is required to put a Right Line at

the Point A, equal to the given Right Line B C.

B 3

Draw

Poft. 1.

I Poft. 2.

* Poft. 3.

Draw the Right Line A C from the Point A to C*, + 1 of this. upon it describe the Equilateral Triangle DAC+; produce D A and DC directly forwards to E and G‡; about the Centre C, with the Distance B C, describe the Circle BGH *; and about the Centre D, with the Distance DG, describe the Circle K GL.

I Ax. 3.

Ax. 1.

Now because the Point C is the Centre of the Circle + Def. 15. BGH, BC will be equal to CG+; and because D is the Centre of the Circle KGL, the Whole DL will be equal to the Whole DG, the Parts whereof DA and DC are equal; therefore the Remainders AL, CG, are also equal ‡. But it has been demonstrated, that BC is equal to CG; wherefore both AL and BC are each of them equal to CG. But Things that are equal to one and the same Thing, are equal to one another *; and therefore likewise A L is equal to B C.

Whence, the Right Line AL is put at the given Point A, equal to the given Right Line BC; which was to be done.

PROPOSITION III.

PROBLEM.

Two unequal Right Lines being given, to cut off a
Part from the greater, equal to the leffer.

LET AB and C be the two unequal Right Lines
given, the greater whereof is AB; it is required
to cut off a Line from the greater A B equal to the
lefler C.

Put *

* 2 of this.

+ Pot. 3.

a Right Line AD at the Point A, equal to the Line C; and about the Centre A, with the Diftance A D, describe a Circle D E F +.

Ax. 1.

Then because A is the Centre of the Circle DEF, A E is equal to AD; and so both AE and Care each equal to A D, therefore A E is likewise equal to C .

And fo, there is cut off from AB the greater of two given Right Lines AB and C, a Line A E equal to the

leffor Line C; which was to be done.

PROPROPOSITION IV.

THEOREM.

If there are two Triangles that have two Sides of the one equal to two Sides of the other, each te each, and the Angle contained by those equal Sides in one Triangle equal to the Angle contained by the correspondent Sides in the other Triangle; then the Base of one of the Triangles shall be equal to the Base of the other, the whole Triangle equal to the whole Triangle, and the remaining Angles of one equal to the remaining Angles of the other, each to each, which fubtend the equal Sides. ET the two Triangles be ABC, DEF, which

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have two Sides A B, AC, equal to two Sides DE, DF, each to each, that is, the Side A B equal to the Side D E, and the Side AC to DF; and the Angle B A C equal to the Angle EDF. I say, that the Bafe BC is equal to the Base EF, the Triangle A B C equal to the Triangle DEF, and the remaining Angles of the one equal to the remaining Angles of the other, each to its correspondent, subtending the equal Sides; viz. the Angle ABC equal to the Angle DEF, and the Angle ACB equal to the Angle DFE.

For the Triangle ABC being applied to DEF, fo as the Point A may co-incide with D, and the Right Line A B with DE, then the Point B will co-incide with the Point E, because A B is equal to DE. And fince A B co-incides with D E, the Right Line AC likewise will co-incide with the Right Line DF, because the Angle BAC is equal to the Angle EDF. Wherefore also C will co-incide with F, because the Right Line A C is equal to the Right Line DF. But the Point B co-incides with E, and therefore the Base BC co-incides with the Base EF. For, if the Point B co-inciding with E, and C with F, the Base BC does not co-incide with the Base EF; then two Right Lines wiil contain a Space, which is impossible *.* Ax. 10. Therefore, the Base BC co-incides with the Base E F, and is equal thereto; and confequently the whole TriB 4 angle