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. Let the st. line AB be divided into any two parts in C. 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. a Let the st. line AB be divided into any two parts in C. Then must sq. on AB=sum of sqq. on AC, CB and Iwice rect. AC, CB. On AB describe the sq. ADEB. 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. Ι. .: BK, KF, FC, CB are all equal ; and KBC is a rt. Z; .: 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, and AE=sq. on AB, HG=sq. on AC, Q. E. D. .. sq. on AB= 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. 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. Let the st. line AB be divided equally in C and unequally in D. Then must rect. AD, together with sq. on CD=sq. on CB. I. 46. On CB describe the sq. CEFB. Draw DG || to CE, and from it cut off DH=DB. Draw HLK || to AD, and AK || to DH. I. 31. I. 31. Then rect. DF=rect. AL, Also LG=sq. on CD, ::BF=AC, and BD=CL. Then rect. AD, DB together with sq. on CD =AH together with LG =sq. on CB. Q. L. D. PROPOSITION VI. THEOREM. If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced. Let the st. line AB be bisected in C and produced to D. sq. I. 46. I. 31. Through H draw KLM || to AD I. 31. Through A draw AK || to CE. Now ::: BG=CD and BH=BD; .. HG=CB; Ax. 3. II. A. Then rect. AD, DB together with sq. on CB =sum of AM and LG sq. on CD. Q. E. D. NOTE. We here give the proof of an important theorem. which is usually placed as a corollary to Proposition V. PROPOSITION B. THEOREM. The difference between the squares on any two straight lines 28 equal to the rectangle contained by the sum and difference of those es. D B K E G Produce AD to B, making CB=AC. and DB=the difference of the lines AC, CD. Then must difference between sqq. on AC, CD=rect. AD, DB. On CB describe the sq. I. 46. Draw DG 11 to CE, and from it cut off DH=DB. I. 31. Draw HLK || to AD, and AK || to DH. I. 31. Then rect. DF=rect. AL, ::BF=AC, and BD=CL. Also LG=sq. on CD, :::LH=CD, and HG=CD. =difference between sqq. on CB, CD Q. E. D. Ex. Shew that Propositions V. and VI. might be deduced from this Proposition. |