Book I. XV. called the circumference, and is such that all straight XVI. XVII. A diameter of a circle is a straight line drawn through the See N. centre, and terminated both ways by the circumference. XVIII. A semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter. XIX. " A segment of a circle is the figure contained by a straight “ line, and the circumference it cuts off.” XX. XXI. XXII. XXIII. Multilateral figures, or polygons, by more than four straight lines. XXIV. XXV. Boor I. ДДД XXVI. XXVII. XXVIII. XXIX. XXX. equal, and all its angles right angles. XXXI. XXXII. are not right angles, a XXXIII. See N. A rhomboid, is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles. Boox I. XXXIV. XXXV. which, being produced ever so far both ways, do not meet. POSTULATES. I. Let it be granted that a straight line may be drawn from any one point to any other point. II. III. distance from that centre. A X IOMS. 1. II. III. IV. V. VI. other. VII. Things which are halves of the same, are equal to one an- VIII. exactly fill the same space, are equal to one another. Book I. IX. X. XI. XII. gether less than two right angles, these straight lines being continually produced, shall at length meet upon 6 that side on which are the angles which are less than “ two right angles. See the notes on Prop. 29. of Book I." Book ! late. Book I. PROPOSITION I. PROBLEM. To describe an equilateral triangle upon a given finite straight line. Let AB be the given straight line; it is required to describe an equilateral triangle upon it. From the centre A, at the distance AB, describes the circle BCD, and from the *3 Postucentre B, at the distance BA, describe the circle ACE; and ! A! B E from the point C, in which the circles cut one another, draw the straight lines CA, i Post. CB, to the points A,B; ABC shall be an equilateral triangle. Because the point A is the centre of the circle BCD, AC is equal to AB; and because the point B is the centre < 15 Definiof the circle ACE, BC is equal to BA : But it has been tion. proved that CA is equal to AB; therefore CA, CB, are each of them equal to AB; but things which are equal to the same are equal to one anotherd; therefore CA is equal a 1st Axion. to CB; wherefore CA, AB, BC are equal to one another; and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB. Which was required to be done. PROP. II. PROB. From a given point to draw a straight line equal. to a given straight line. Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC. K From the point A to B drawa * 1 Post. the straight line AB; and upon it describe the equilateral triangle H DAB, and produce the straight D • 2 Post. lines DA, DB, to E and F; from A the centre B, at the distance BC BI described the circle CGH, and d3 Post. from the centre D, at the distance E DG describe the circle GKL. AL shall be equal to BC. b 1. 1. |