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 bath three unequal Sides. XXVII. Also of three-fided Figures, that is a Right-angled Triangle, which bath 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; buti 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. POSTULATE S. IGRANT RANT that a Right Line may be drawn from any one Point to another. II. That a finite Right Line may be continued direally forwards. III. And that a Circle may be defcribed about any Centre with any Distance. B 2 AXIOMS. AXIOM S. THINGS 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 Part. X. Two Right Lines do not contain a Space. XI. All Right Angles are equal between themfelves. 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 A B, 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. To defcribe an Equilateral Triangle upon a given finite Right Line. L ET AB be the given finite Right Line, upon About the Centre A, with the Distance A B, defcribe the Circle BCD*; and about the Centre B,* Post. 3. with the fame Diftance BA, defcribe the Circle ACE*; and from the Point C, where the two Circles cut each other, draw the Right Lines CA, CB+. † Poft. I. Then because A is the Centre of the Circle D BC, A Chall be equal to A B‡. And becaufe B is the ‡ Def. 15. Centre of the Circle CA E, BC fhall 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 themfelves, and confequently AC is* Ax, 1. equal to CB; therefore the three Sides CA, AB, BC, are equal between themfelves. And fo, the Triangle BAC is an Equilateral one, and is defcribed 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 LET the Point given be A, and the given Right B 3 Draw Poft. 1. Pot. 2. 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 Centre C, with the Distance B C, defcribe the Circle BGH*; and about the Centre D, with the Distance DG, defcribe the Circle K GL. Poft. 3. 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 K GL, the Whole DL will be equal to the Whole DG, the Parts whereof DA and DC are equal; therefore the Remainders Ax. 3. AL, CG, are alfo equal t. But it has been demonftrated, that BC is equal to CG; wherefore both A L and BC are each of them equal to C G. But Things that are equal to one and the fame Thing, are equal to one another * ; and therefore likewife A L is equal to B C. #Ax. 1. Whence, the Right Line AL is put at the given Point A, equal to the given Right Line BC; which was to be done. 2 of this. PROPOSITION III. PROBLEM. Two unequal Right Lines being given, to cut off a LE ET AB and C be the two unequal Right Lines given, the greater whereof is A B; 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 Centre A, with the Dif Po stance A D, defcribe a Circle D E F †. Ax. 1. Then because A is the Centre of the Circle DEF, AE is equal to AD; and fo both AE and C are each equal to A D therefore AE is likewife equal to CT. And fo, there is cut off from AB the greater of two given Right Lines A B and C, a Line A E equal to the leffr 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 fhall 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. LE ET the two Triangles be A B C, DEF, which have two Sides A B, A C, 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 fay, that the Bafe B C is equal to the Bafe EF, the Triangle A B C equal to the Triangle D E F, 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 A C B 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 D E, then the Point B will co-incide with the Point E, because A B is equal to D E. And fince A B co-incides with D E, the Right Line A C likewife will co-incide with the Right Line DF, because the Angle BAC is equal to the Angle E DF. Wherefore alfo 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 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 wiil contain a Space, which is impoffible Therefore, the Bafe B C co-incides with the Bafe E F, and is equal thereto; and confequently the whole Tri B 4 angle * Ax. 10. |