PROPOSITION I. THEOREM. If there be two straight lines, one of which is divided into any number of parts, 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. B H Let AB and CD be two given st. lines, and let CD be divided into any parts in E, F. Then must rect. AB, CD=sum of rect. AB, CE and rect. AB, EF and rect. AB, FD. From C draw CG 1 to CD, and in CG make CH=A.B. Through E, F, and D draw EK, FL, DM || to CH. I. 31. .. rect. AB, CD=sum of rect. AB, CE and rect. AB, EF and rect. AB, FD. Q. E. D. Ex. If two straight lines be each divided into any number of parts, the rectangle contained by the two lines is equal to the rectangles contained by all the parts of the one taken separately with all the parts of the other. PROPOSITION II. THEOREM. If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts are together equal to the square on the whole line. Let the st. line AB be divided into any two parts in C. Then must sq. on AB=sum of rect. AB, AC and rect. AB, CB. .. sq. on AB=sum cf rect. AB, AC and rect. AB, CB. Q. E. D. Ex. The square on a straight line is equal to four times the square on half the line. PROPOSITION III. THEOREM. If a straight line be divided into any two parts, 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 on the aforesaid part. B Let the st. line AB be divided into any two parts in C. Then must rect. AB, CB=sum of rect. AC, CB and sq. on CB. On CB describe the sq. CDEB. I. 46. From A draw AF || to CD, meeting ED produced in F. Then AE sum of AD and CE. Now AE-rect. AB, CB, ::: BE=CB, AD=rect. AC, CB, :: CD=CB, CE=sq. on CB. .. rect. AB, CB sum of rect. AC, CB and sq. cn CB. Q. E. D. NOTE. When a straight line is cut in a point, the distances of the point of section from the ends of the line are called the segments of the line. If a line AB be divided in C, AC and CB are called the internal segments of AB. If a line AC be produced to B, AB and CB are called the external segments of AC. PROPOSITION IV. THEOREM. If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts together with twice the rectangle contained by the parts. Let the st. line AB be divided into any two parts in C. sq. on AB=sum of sqq. on AC, CB and twice rect. AC, CB. I. 46. From AD cut off AH=CB. Then HD=AC. Draw CG || to AD, and HK || to AB, meeting CG in F. Then. BK=AH, .. BK=CB, Ax. I. .. BK, KF, FC, CB are all equal; and KBC is a rt. ≤ ; .. CK is the sq. on CB. Def. xxx. Also HG=sq. on AC, · HF and HD each=AC. Now AE sum of HG, CK, AF, FE, .. sq. on AB=sum of sqq. on AC, CB and twice rect. AC, CB. Q. E. D. Ex. In a triangle, whose vertical angle is a right angle, a straight line is drawn from the vertex perpendicular to the base. Shew that the rectangle, contained by the segments of the base, is equal to the square on the perpendicular. S.E. PROPOSITION V. THEOREM. If a straight line be divided into two equal parts and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line. H L E Let the st. line AB be divided equally in C and unequally in D. Then must rect. AD, DB together with sq. on CD=sq. on CB. On CB describe the sq. CEFB. I. 46. Draw DG to CE, and from it cut off DHDB. I. 31. Then rect. AD, DB together with sq. on CD =AH together with LG =sum of AL and CH and LG =sum of DF and CH and LG = CF = =sq. on CB. Q. E. D. |