A circle is a plain figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another. XVI. XVII. A diameter of a circle is a straight line drawn thro' the center, See N. 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 fegment of a circle is the figure contained by a straight line « and the circumference it cuts off.” XX. XXI. XXII. XXIII. XXIV. XXV. Book I. AAA XXIV XXVI. XXVII. XXVIII. AAN Ҳу. XXIX. XXX. equal, and all its angles right angles. XXXI. XXXII. are not right angles. 77 See N. XXXIII. ther, but all its fides are not equal, nor its angles right angles. Book I. XXXIV. XXXV. which, being produced ever so far both ways, do not meet. POSTULAT E S. L I. II. III. distance from that center. A X I O M S Tanother . I. II. III. IV. V. VI. VII. VIII. Took 1. IX. X. XI. XII. “ two interior angles on the fame side of it taken together less « than two right angles, these straight lines being continually “ produced shall at length meet upon that fide on which are “ the angles which are less than two right angles. See the notes on Prop.-29. of Book I.” Book I. finite straight line. it. From the center A, at the distance AB describe a the circle a. 3d Postue late. BCD. and from the center B, at the distance BA describe the circle D A . B E ACE; and from the point C in which the circles cut one another draw the straight lines b CA, CB to the points A, B. ABC shall be an equilateral triangle. Because the point A is the center of the circle BCD, AC is equal e to AB. and because the point B is the center of the circle ACE, C. 15th De. finition. BC is equal to BA. but it has been 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 another d; therefore d. ift Axie CA is equal to CB. wherefore CA, AB, BC are equal to one another. and the triangle ABC is therefore equilateral, and it is defcribed upon the given straight line AB. Which was required to be done. b. 2d Poft. om. PRO P. II. PRO B. to a given straight line. a. I. Poft. Straight line AB; and upon it de K scribe the equilateral triangle DAB, H and produce the straight lines DA, DB to E and F; from the center B, at the distance BC describe d the d. 3. Pott circle CGH, and from the center D, B В at the distance DG described the G E circle GKL. AL Thall be equal to BC. b. I. I. c. 2. Poft, |