XXIV. Of tbree-sided Figures tbat is an Equi lateral Triangle, which baib ibree. equal Sides. XXV. That an Ijosceles, or Equicrural one, wbich bath only two Sides equal. XXVI. And a Scalene one, is that, which hatb three unequal Sides. XXVII. Also of obrce-sided Figures, that is a Right-angled Triangle, which baih a Right Angle. XXVIII. That an Obtuse-angled one, which bath an Obtufe Angle. XXIX. And ibat an Acute-angled one, which bath three Acute Angles. XXX. Of four-sided Figures, that is a Square, whose four Sides are equal, and its Angles all Rigbt ones. XXXI. That an Oblong, or Rectangle, which is longer than bro.d; buti its oppofile Sides are equal, and all its Angles Right ones. XXXII. That a Rbombus, which bath four equal Sides, but not Right Angles. XXXIII. That a Rbomboides, whose opposite Sides and Angles only are equal. XXXIV. All Quadrilateral Figures, besides these, are called Trapezia, XXXV. Parallels are such Right Lines, in the Same Plane, which, if infinitely produc'd burb Ways, would never meet. P O S T U L A T E S. from any one point to another. II. That a finite Right Line may be continued di really forwards. III. And that a Circle may be described about any Centre with any Distance. B 2 AXIOM S. M A X I O M S. 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 Wboles 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 same 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. Iwo Right Lines do not contain a Space. XI. All Right Angles are equal between them felves. XII. If a Right Line, falling upon two other Right Lines, makes the inward Angles on the Joine Side tkereof, 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 ihree 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. PROPROPOSITION I. PROBLEM. To describe an Equilateral Triangle upon a given finite Right Line. L ET AB be the given finite Right Line, upon Triangle. About the Centre A, with the Distance A B, defcribe the Circle BCD*; and about the Centre B, * Poff. 30 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, CBt. Poft. 1. Then because A is the Centre of the Circle D BC, AC Ihall be equal to A BI. And because B is the t Def. 15. Centre of the Circle CAE, BC shall be equal to BA: But C A hath been proved to be equal to AB; therefore both CA and C B are each equal to A B. But Things equal to one and the fame Thing, are equal between themselves *, and consequently A Cis* Ax, 1. equal to CB; therefore the three Sides CA, AB, B C, are equal between themselves. And so, 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. Line BC; it is required to put a Right Line at the Point A, equal co the given Right Line B C. B 3 Draw * Poft. 3. Pop. 1. Draw the Right Line AC from the Point A to C*, +1 of this. upon it describe the Equilateral Triangle DAC+ i Pol. 2. produce D A and DC directly forwards to E and Gf; about the Centre C, with the Distance BC, describe the Circle B GH*; and about the Centre D, with the Distance DG, describe the Circle KGL. Now because the Point C is the Centre of the Circle + Def. 15. BGH, BC will be equal to CGt; 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 'D C are equal; therefore the Remainders | Ax. 3. AL, CG, are also equal I. But it has been demon strated, that B C 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 * ; nd therefore likewise A L is equal to B C. Whence, the Right Line A L is put at the given Point A, equa! 10 the given Right Line B C; which was to be done. Ax. I. PROPOSITION III. PROBLEM. Two unequal Right Lines being given, to cut off a Part from tbe greater, equal to the lefer. L. 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 leller C. * 2 of this. Put * a Right Line A D at the Point A, equal to the Line C; and about the Centre A, with the Dir† Po 4. z. tance AD, describe a Circle D E F t. Then because A is the Centre of the Circle DEF, AE is equal to A D; and so both A E and Care each equal to AD; therefore A E is likewise equal Ax. ?. to CI. And so, 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 lelj ir Line C; which was to be done. PROPROPOSITION IV. THEOREM. the one equal to two sides of the other, each to other each to each, which subtend the equal Sides. LE ET the two Triangles be ABC, 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 DE, and the Side A C to DF; and the Angle B A C equal to the Angle EDF. I say, that the Base B C is equal to the Base E F, the Triangle A B C equal to the Triangle D EF, and the remaining Angles of the one equal to the remaining Angles of the other, each to its correspondent, fubtending the equal Sides ; viz. the Angle A B C equal to the Angle D E F, and the Angle & C B equal to the Angle DFE. For the Triangle A B C being applied to DEF, so as the Point A may co-incide with D, and the Right Line A B with DÉ, 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 AC likewise will co-incide with the Right Line DF, because the Angle BAC is equal to the Angle EDF. Wherefore allo C will co-incide with F, because the Right Line AC 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 E F; then two Right Lines wiil contain a Space, which is impoffible *.* Ax. 16. Therefore, the Base B C co-incides with the Bafe E F, and is equal thereto ; and consequently the wbole Tri angle B4 |