nearer to the extremity B, the rectangle under AC and CB continually diminishes; and when C is indefinitely near to B, the rectangle is indefinitely small, approaching to evanescence; and the sum of the squares on the parts, FE and CK, then differs from the entire square by a quantity indefinitely little. The square on the line is thus the limit towards which the sum of the squares on the parts continually approaches, as the rectangle diminishes towards evanescence, and to which we can make the sum approach as near as we please. Cor. 4.-Since the difference of the squares on the sides of a triangle is equal to the difference of the squares on the segments of the base, made by a perpendicular from the vertex (I. 37, cor. 3), it follows from Cor. 1 to this Proposition, that the rectangle under the sum and difference of the sides is equal to the rectangle under the sum and difference of the segments of the base; that is, to the rectangle under the base and the difference of its segments, or to twice the rectangle under the base, and half the difference of its segments. Now, this half difference is the distance of the foot of the perpendicular from the middle of the base, for D being the middle point, if DK be laid off equal to DO, CK is equal to BO (ax. 3), and the difference of CO and OB is OK; that is, OD is half the difference of the segments. Thus the rectangle under CB and OK is the same as that under twice CB and OD. Thus the rectangle under the sum and difference of the sides is equal to twice the rectangle under the base, and half the difference of the segments. But if the perpendicular fall outside the base, the base BC is the difference of the segments CO and OB (I. 37, cor. 3, b). Now, CO is equal to CB and BO, and therefore CO and OB are equal to CB and twice BO; therefore also CO and OB are equal to twice DB and twice BO, since D is the middle point of CB; therefore the half of CO and OB is equal tô DB and BO; that is, DO is half the sum of the segments. The distance of the foot of the perpendicular from the middle of the base is thus half the sum of the segments when the perpendicular falls outside the base; and the rectangle of the sum and difference of the sides is equal to twice the rectangle under the base and half the sum of the segments. Cor. 5.-Since the square on one side of a right-angled triangle is equal to the excess of the square on the hypotenuse above the square on the other side (I. 37, cor. 2), it follows that the square on one of the sides is equal to the rectangle under the sum and difference of the hypotenuse and the other side. Cor. 6.-a. The line AC is (const.) equal to half the sum of AD and DB; and if CK be laid off towards A, equal to CD, it will appear that CD is half the difference of AD and DB, or AD and AK (ax. 3). If then the half difference CD be added to the half sum AC, we shall have the greater line AD; and if the half difference be taken from the half sum, we shall have the less line DB; that is, the greater is half the sum of the sum and difference, and the less half the difference of the sum and difference. b. Also, since C is the middle point, it is evident that AC and CD are equal to DB and twice CD; that is, that the sum of two lines is equal to twice the less together with the difference. c. Again, since CD is half the difference of the sides of the rectangle, this Proposition may be thus stated: "The rectangle under any two lines, together with the square on half their difference, is equal to the square on half their sum." d. The lines AD, AC, DB are plainly three equidifferent quantities, CD being the common difference; and the middle term or mean, AC, is half the sum of the extremes AD and DB. Cor. 7.-If two equal straight lines be so divided, that the rectangle under the segments of G E B lines be bisected, the thing is evident; but if not, let E and F be the points of division, and let the lines be bisected in G and H. Then the rectangle under AE and EB is equal to that under CF and FD (hyp.); and if these equals be taken from the equals, the squares on half the lines (I. 36, cor. 1), the remainders, the squares on GE and HF, are equal (II. 5, ax. 3); therefore GE is equal to HF (I. 36, cor. 2); therefore the sums AE and CF are equal (ax. 2), and also the differences EB and FD. Cor. 8.-The rectangle under AD and DB is bounded by the same extent of line as the square on BC, that is, the figures have the same perimeter, or are isoperimetrical; and since the square has a greater area by the square on the intercept CD, it follows that the area of the square is greater than that of any other isoperimetrical figure; and if a square and a rectangle have the same area, the perimeter of the square is less than that of the rectangle. PROP. VI.-THEOREM. (EUC. II. 6.) If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced, and the part produced, together with the square on half the line bisected, is equal to the square on the straight line made up of the half and the part produced. K F Ꮮ Let the straight line AB be bisected in C and produced to D, the rectangle contained by the whole line AD, and the part DB, together with the square on N CB, is equal to the square on CD. From D draw DE perpendicular to AD (I. 7, cor. 1), making it equal to DB, and construct the rectangle AE. M D On CB describe the square E CBGF (I. 36), and on CD the Square CDLK. Produce FC through C, meeting VE in M; and FG through G, meeting DL in N. Then, as before, the rectangle AM is equal to the rectangle BN, BN and NK together. Add to both FB, which is the square on CB; therefore the rectangle AE, together with the square on CB, is equal to the whole figure CL, which is the square on CD. together with the square on CB, is equal to the square on CD. Q. E. D. (Ax. 2.) If a straight line be divided into any two parts, the square on the whole line, together with the square on one of the parts, is equal to twice the rectangle contained by the whole line and that part, together with the square on the other part. Let AB be a straight line divided into any two parts in C; it is required to prove that the square on the whole line AB, On AB describe the square ADEB (I. 36), on CB describe the square CBKH, and on AC the square AFGC; and let these squares stand as in the figure. F D Produce FG through G, meeting BE in L. H K B Then EL is equal to BC, and AC to BL; (Ax. 3.) also LK is equal to AB. Now, the two squares AE and CK together are equal to the square AG, (Ax. 2.) together with the two rectangles DL and LH. (Ax. 10.) But AE is the square on AB, DL the rectangle contained by AB and BC, since EL is equal to BC; and LH is also the rectangle contained by AB and BC, since LK is equal to AB. Therefore the square on the whole line AB, together with the square on the part BC, is equal to twice the rectangle contained by AB and BC, together with the square on the other part AC. Q. E. D. Schol. Since AC is the difference between AB and BC, the proposition may be thus stated: "The sum of the squares on two straight lines exceeds the square on their difference by twice the rectangle of the two lines." Cor.-Since the square on the sum of two lines exceeds the sum of their squares by twice their rectangle, and the sum of the squares the square on the difference by twice the rectangle (schol. preceding), it follows that the square on the sum exceeds the square on the difference by four times the rectangle. Hence, also, the square on the sum, the sum of the squares, and the square on the difference, are three equidifferent quantities, the common difference being twice the rectangle under the two lines. PROP. VIII.—THEOREM. (EUC. II. 8.) 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 that part. C B D Let the straight line AB be divided into any two parts in C, then shall four times the rectangle contained by AB and BC, together with the square on AC, be equal to the square on the line which is made up of AB and BC together. H K Produce AB to D, making BD equal to BC; then AD is made up of AB and BC together. On AD describe the square ADFE (I. 36); from B draw BG perpendicular to AB (I. 7, cor. 1), and make BG equal to BC. Through G draw GH parallel to AB (I. 22), meeting AE |