From the point A draw AD at right angles to AC, (1. 11) and make AD equal to BA, (1. 3) and join DC. DEMONSTRATION Then, because DA is equal to AB, the square of DA is equal to the square of AB; to each of these equals add the square of AC, therefore the squares of DA, AC, are equal to the squares of BA, AC, (ax. 2) but because DAC is a right angle, (constr.) therefore the square of DC is equal to the squares of DA, AC; (1. 47) and the square of BC is equal to the squares of BA, AC; (hyp.) therefore the square of DC is equal to the square of BC; (ax. 1) wherefore also the side DC is equal to the side BC. Hence, in the two triangles DAC, BAC, because the side DA is equal to the side AB, (constr.) and AC common to both triangles, the two DA, AC, are equal to the two BA, AC, each to each, and the base DC has been proved to be equal to the base BC; therefore the angle DAC is equal to the angle BAC; (1. 8) but DAC is a right angle; (constr.) therefore also BAC is a right angle. (ax. 1.) Therefore, if the square, &c. Q. E. D. 59 BOOK II. DEFINITIONS. I. ' EVERY right-angled parallelogram, or rectangle, is said to be contained by any two of the straight lines which contain one of the right angles. II. In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a Gnomon. 'Thus the parallelogram HG, together with the complements AF, FC, is the Gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which make the Gnomon.' PROP. I.-THEOREM. If there be two straight lines, one of which is divided into any number of parts; then the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line. (References-Prop. I. 3, 11, 31, 34.) Let A and BC be two straight lines; and let BC be divided into any number of parts in D and E. Then the rectangle contained by the straight lines A and BC is equal to the rectangle contained by A and BD, together with that contained by A and DE, and A and EC. From the point B, draw BF at right angles to BC, (1. 11) and make BG equal to A; (1. 3) through G draw GH parallel to BC; (1. 31) and through the points D, E, C, draw DK, EL, CH, parallel to BG. DEMONSTRATION Then the rectangle BH is equal to the rectangles BK, DL, EH; but because BH is contained by GB, BC, and GB is equal to A; (constr.) therefore BH is contained by A and BC, and, because BK is contained by GB, BD, of which GB is equal to A, therefore BK is contained by A and BD; and because DL is contained by DK and DE, and DK, that is, BG, (1. 34) is equal to A; therefore DL is contained by A and DE; and in like manner EH is contained by A and EC; therefore the rectangle contained by A and BC is equal to the several rectangles contained by A and BD, by A and DE, and by A and EC. Wherefore, if there be two straight lines, &c. Q E. D. PROP. II.-THEOREM. If a straight line be divided into any two parts; then the rectangles contained by the whole and each of the parts are together equal to the square of the whole line. Let AB be divided into any two parts in C. Then the rectangle* contained by AB and BC, together with the rectangle contained by AB and AC, shall be equal to the square of AB. Upon AB describe the square ADEB, (1. 46) DEMONSTRATION Then AE is equal to the rectangles AF, CE. But AE is the square of AB; therefore the square of AB is equal to the rectangles AF, CE. But AF is contained by DA, AC, of which AD is equal to AB, (def. 30) therefore AF is the rectangle contained by AB, AC; and CE is contained by AB, BC, since BE is equal to AB; therefore the rectangle contained by AB, AC, together with the rectangle AB, BC, is equal to the square of AB. If, therefore, a straight line, &c. Q. E. D. PROP. III.-THEOREM. If a straight line be divided into any two parts; then the rectangle contained by the whole and one of the parts is equal to the rectangle contained by the two parts, together with the square of the aforesaid part. (References-Prop. I. 31, 46.) Let the straight line AB be divided into any two parts in the point C. N.B. To avoid repeating the word contained too frequently, the rectangle contained by two straight lines, AB and AC, is sometimes simply called the rectangle AB, AC. Then the rectangle AB, BC, shall be equal to the rectangle AC, CB, together with the square of CB. Upon BC describe the square CDEB; (1. 46) produce ED to F; and through A draw AF parallel to CD or BE. (I. 31.) DEMONSTRATION Then the rectangle AE is equal to the rectangles AD, CE. But AE is the rectangle contained by AB, BE, of which BE is equal to BC, (def. 30) therefore AE is the rectangle contained by AB, BC; and AD is contained by AC, CD, of which CD is equal to BC, therefore AD is contained by AC, CB; and CE is the square of BC; therefore the rectangle AB, BC, is equal to the rectangle AC, CB, together with the square of BC. If, therefore, a straight line be divided, &c. Q. E. D. PROP. IV.-THEOREM. If a straight line be divided into any two parts; then the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts. (References Prop. 1. 5, 6, 29, 31, 34, 43, 46.) Let the straight line AB be divided into any two parts in C. |