square = twice Since the line is divided into two equal parts, the rectangle contained by the two parts will be the same as the on half the line, hence the 'squares on the two parts' the square on half the line, and twice the rectangle contained by the parts =twice the square on half the line; that is, the square on the whole line = four times the square on half the line. SUMMARY OF PROPOSITION IV., THEOREM 4. 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 two parts. (Sq. on AB sqq. on AC, CB, and twice rect. AC, CB.) Cons. Draw a line through the pt. of section parallel to the sides of the square on the whole line. Proof. Each of the rects. into which the sq. on AB is divided = the square on one of the parts and rect. AC, CB (II. 3). ..the 2 rects. together, that is the square on AB, squares on the two parts and twice rect. AC, CB. = the 2ND METHOD. Cons. From BE cut off BG = BC. Through pts. C and G draw lines | to opposite sides of sq. on AB. Proof.-AE, that is sq. on AB, sum of CG and KF = sqq. on AC, CB), and rects. AH, HE (= twice rect. AC, CB). 3RD (EUCLID'S) METHOD. Cons.-Draw CF || AD, join BD. Through pt. H, at which these lines intersect, draw GHK .. since CG is a rect. parallelogram by cons. it is a square (I. 34). For the same reason KF is a sq. on KH = AC (I. 34). .. four figures KF, CG, AH, and HE = squares on AC, CB and twice rect. AC, CB, and together make up sq. on AB. -Q.E.D. LECTURE XIX. PROPOSITIONS 5, 6, AND 8. We have seen that the first three propositions may be included under one enunciation. Props. 5, 6, and 8 form a second group of three propositions, each of which is a particular case of the general law, that 'the rectangle contained by two straight lines, together with the square on half their difference, is equal to the square on half their sum.' : To make this clear let us take a concrete example :Given two straight lines, 6 in. and 2 in. long. Area of the rectangle contained by the two straight lines = 6 × 2: = 12 sq. in. = Area of square on half their difference sq. in. i.e. area of rectangle, + square on half the difference=16 sq. in.; area of square on half their sum = sq. in. () We proceed to the geometrical proof of this law. and DB be the two given straight lines, and let the same straight line with AD. A C = 42 = 16 Let AD DB be in K L H Then the line AB is the sum of AD and DB, bisect it in C, and on CB describe a square CBFE (I. 46). This is the square on half of their sum. To describe the rectangle ADDB, draw DG parallel to CE or BF, one of the opposite sides of the square, and cut off from it DH equal to DB. AD and DH will be the sides containing a right angle; complete the rectangle as in former propositions. E G F It remains to construct the square on half the difference of the two straight lines. Now AD the difference + DB. = AB = AD + DB = the difference + 2 DB. Therefore half AB, that is CB = half the difference and DB, and CB = CD + DB. Therefore CD = half of the difference of the two lines. From the construction in the last proposition (II. 4), we know that LG is a square, and its side LH = = CD. Then LG is the square on half the difference of AD and DB; and CF, the square on half the sum, is equal to LG, the square of half the difference, together with the rectangle CH, a part of rectangle AD, DB, and the rectangle DF. Is the rectangle DF equal to AL the remaining part of AH? GF = DB, and BF = CB (I. 46) = AC. Therefore the rectangle AL = DF. Consequently CF, the square on half the line, the rectangle AD, DB (AH), together with the square on CD (LG). In constructing the figure for this demonstration, we took care to place AD and DB in one straight line, in order to obtain a measure of their sum; and the whole line AB was bisected at C, and divided into two unequal parts at D Hence, supposing one line AB to be given, the proposition may be enunciated : If a straight line (AB) be divided equally (at C) and unequally (at D), the rectangle contained by the unequal parts (ADDB), together with the square on the line between the points of section (CD), is equal to the square on half the line. This is the form of enunciation which Euclid adopts in the fifth proposition; his demonstration differs slightly from that given above, for having described the square on CB, he draws its diameter and adopts the construction followed in II. 4. Proposition VI. presents the second case of the general enunciation given above. If a straight line AB be bisected in C, and produced to any point D, the rectangle contained by the whole line thus produced, and the part of it produced (rect. AD, DB), together with the square on half the line bisected (CB), is equal to the square on (CD) the straight line, which is made up of the half and the part produced. That this is a particular case of the enunciation referred to will easily be seen. Let AD and DB be the A B D given lines; then AB is their difference. Since AB is bisected in C, CB is half their difference. Sum of the two lines the difference + 2 BD. = CD. The construction and proof are similar to that given, making the necessary change in constructing the rectangle ADDB. If Euclid adopts the same construction as before. We pass to Proposition VIII., the third of the group. a straight line AB be divided at C into any two parts, four times the rectangle contained by the whole line and one of the parts (4 AB, BC), together with the square on AC, the other part, is equal to the square on the straight line, which is made up of the whole and that part (AB produced to D, so that BD = BC). Again, supposing AB, BC to be two given lines, we may state the enunciation thus :— Four times the rectangle contained by the two lines, together with the square on their difference (AC), is equal to the square on their sum (AB + BC). From the corollary to II. 4, we learn that the square on any line is equal to four times the square on half that line. Whence we obtain this equation : Four times the rectangle contained by the two lines, together with four times the square on half their difference = four times the square on half their sum. Hence it is clear that this proposition also may be inferred from the enunciation which we have given of II. 5. It may also easily be proved as an independent demonstration by either of the methods indicated in the last three propositions. |