ร

XXIV. Of three-fided Figures, that is, an Equilateral Triangle, which bath three equal Sides. XXV. That an Ifofceles, 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-fided Figures, that is, a
Right-angled Triangle, which hath a Right
Angle.

XXVIII. That an Obtufe-angled one, which bath
an Obtufe Angle.

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

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

XXXI. That an Oblong, or Rectangle, which is longer than broad; but its oppofite 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, whofe oppofite Sides and Angles only are equal.

XXXIV. All Quadrilateral Figures, befides thefe, are called Trapezia.

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

I.

G

POSTULATES.

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

II. That a finite Right Line may be continued di-
rectly forwards.

III. And that a Circle may be described about any Center, with any Distance.

TH

AXIOM S.

I. HINGS equal to one and the fame
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 fame 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 Parts. X. Two Right Lines do not contain a Space. XI. All Right Angles are equal between themselves. 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, thofe 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 feveral Angles at one Point, any one of them is exprefs'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 contain'd under the Right Lines AB, B E, is called the Angle ABE.

PRO

PROPOSITION I.

PROBLEM.

To defcribe an Equilateral Trangle 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 Center A, with the Distance AB, defcribe the Circle BCD*; and about the * 3 Poft. Center B, with the fame Distance B A, describe the Circle A CE; and from the Point C, where the two Circles cut each other, draw the Right Lines CA, CB+.

ti Poft.

Then because A is the Center of the Circle DB C, AC fhall be equal to AB. And because B is the 15 Def. Center of the Circle CA E, BC shall be equal to BA: but CA hath been proved to be equal to AB; therefore both CA and CB are each equal to AB. But Things equal to one and the fame Thing, are equal between themselves, and confequently CA is equal to CB; therefore the three Sides CA, A B, B C, are equal between themselves.

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

PROPOSITION II.

PROBLEM.

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

L

ET the Point given be A, and the given Right Line BC; it is required to put a Right Line at the Point A, equal to the given Right Line B C.

* Poft. I. Draw the Right Line A C from the Point A to C*, +1 of this. upon it defcribe the Equilateral Triangle DAC+; produce DA and DC directly forwards to E and G; about the Center C, with the Distance BC, defcribe the Circle BG H*; and about the Center D, with the Distance D G, defcribe the Circle G KL.

Post. 2. *Poft. 3.

+ Def. 15.

Now because the Point C is the Center of the Circle BGH, BC will be equal to CG+; and because D is the Center of the Circle G KL, the whole DL will be equal to the whole D G, the Parts whereof DA and D C are equal; therefore the Remainders Axiom 3. AL, GC are alfo equal t. But it has been demonftrated, 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 fame Thing, are equal to one another; and therefore likewise A L is equal to BC.

* 2 of this.

Poft. 3.

Whence the Right Line AL is put at the given Point A, equal to the given Right Line B C, 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.

ET 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 leffer C.

*

Put a Right Line AD at the Point A, equal to the Line C, and about the Center A, with the Distance AD, defcribe a Circle DEF+.

Then because A is the Center of the Circle DEF, AE is equal to AD; and fo both AE and C are each equal to AD; wherefore AE is likewife equal ‡ Axiom 1. to C‡.

And fo there is cut off from A B the greater of two given Right Lines A B and C, a Line A E equal to the leffer Line C; which was to be done.

PRO

PROPOSITION IV.

THEOREM.

If there are two Triangles that have two Sides of the one equal to two Sides of the other, each to each, and the Angle contained by thofe equal Sides in one Triangle equal to the Angle contained by the correfpondent Sides in the other Triangle, then the Bafe of one of the Triangles shall be equal to the Bafe 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 A B C, DEF, which have two Sides AB, AC, equal to two Sides DE, DF, each to each, that is, the Side A B equal to the Side DE, and the Side AC to DF; and the Angle B A C equal to the Angle EDF. I fay, that the Bafe BC is equal to the Bafe EF, the Triangle ABC equal to the Triangle DEF, and the remaining Angles of the one equal to the remaining Angles of the other, each to its Correfpondent, fubtending the equal Sides, viz. the Angle ABC equal to the Angle DEF, and the Angle ACB equal to the Angle DFE.

For the Triangle A B C 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 DE, the Right Line A C likewife will co-incide with the Right Line D F, because the Angle BAC is equal to the Angle EDF. Wherefore alfo C will co-incide with F, because the Right Line AC is equal to the Right Line D F. But the Point B co-incides with E, and therefore the Base BC co-incides with the Bafe EF. For if the Point B co-inciding with E, and C with F, the Bafe BC does not co-incide with the Base EF; then two Right Lines will contain a Space, which is impoffible *. Therefore the Bafe B C co-incides with the Bafe EF, and is equal thereto; and confequently the whole Triangle

B 3

**Ax.