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. rect. AD, DB together with sq. on CB=sq. on CD. On CD describe the sq. CEFD. I. 46. Draw BG || to CE, and cut off BH=BD. I. 31. Through H draw KLM || to AD I. 31. Through A draw AK || to CE. Now. BGCD and BH=BD; .. HG=CB; .. rect. MG=rect. AL. Then rect. AD, DB together with sq. on CB =sum of AM and LG sum of AL and CM and LG -sum of MG and CM and LG sq. on CD. Ax. 3. II. A. 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 is equal to the rectangle contained by the sum and difference of those lines. Let AC, CD be two st. lines, of which AC is the greater, and let them be placed so as to form one st. line AD. Produce AD to B, making CB=AC. Then AD the sum of the lines AC, CD, 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. CEFB. I. 46. Draw DG 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. Ex. Shew that Propositions V. and VI. might be deduced from this Proposition. PROPOSITION VII. THEOREM. If a straight line be divided into any two parts, the squares on the whole line and on one of the parts are equal to twice the rectangle contained by the whole and that part together with the square on the other part. Let AB be divided into any two parts in C. Then must sqq. on AB, BC=twice rect. AB, BC together with sq. on AC. On AB describe the sq. ADEB. From AD cut off AH=CB. Draw CF to AD and HGK || to AB. Then HF=sq. on AC, and CK=sq. on CB. Then sqq. on AB, BC=sum of AE and CK I. 46. I. 31. .. sqq. on AB, BC=twice rect. AB, BC together with sq. on AC. Q. E. D, Ex. If straight lines be drawn from G to B and from G to D, shew that BGD is a straight line. If a straight line be divided into any two parts, four times the rectangle contained by the whole line and one of the parts, together with the square on the other part, is equal to the square on the straight line which is made up of the whole and the first part. On AD describe the sq. AEFD. H Let the st. line AB be divided into any two parts in C. Produce AB to D, so that BD=BC. Then must four times rect. AB, BC together with sq. on AC=sq. on AD. From AE cut off AM and MX each=CB. I. 46. Through C, B draw CH, BL || to AE. I. 31. Through M, X draw MGKN, XPRO || to AD. I. 31. Now XE AC, and XP=AC, .. XH=sq. on AC. = = four times AK four times rect. AB, BC. Then four times rect. AB, BC and sq. on AC =sum of the eight rectangles and XH =AEFD 8q. on AD. Q. E. D. PROPOSITION IX. THEOREM. If a straight line be divided into two equal, and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section. Let AB be divided equally in C and unequally in D. sum of sqq. on AD, DB=twice sum of sqq. on AC, CD. Draw CE AC at rt. 8 to AB, and join EA, EB. Draw FG at rt. 4s to EC, and join AF. So also BEC and 4 EBC are each-half a rt. 4. L Also, GEF is half a rt. 4, and EGF is a rt. 4; .. LEFG is half art. 4; .. L EFG= 4 GEF, and .. EG=GF. So also 4 BFD is half a rt. 4, and BD=DF. Now sum of sqq. on AD, DB =sq. on AD together with sq. on DF I. B. Cor. sq. on AF =sq. on AE together with sq. on EF I. 47. I. 47. =sqq. on AC, EC together with sqq. on EG, GF I. 47. =twice sq. on AC together with twice sq. on GF Q. E. D. |