THEOREM L. If a straight line be divided into two equal parts, and also into two unequal parts, the square on half the line is equal in area to the rectangle contained by the two unequal parts, together with the square on the part between the points of section. Let A B be the line, bisected in C and divided into unequal parts in D. A C D B G K L F H Let C BEF be the square on CB, and AD L G the rectangle A D, DB; then the square CBEF is equal to the rectangle ADL G, together with the square on CD. Let G L cut C F in K, then C K = DL = DB and KLCD. Continue D L to meet FE in H. Then D H = BE = СВ. Therefore LH = DH-DL CB-DB = CD. Again, since AC = C B = BE and CK = D B, the rectangle A K = the rectangle D E. But the square CBE F is made up of the rectangle CL, the rectangle D E, and the square K H, of which the rectangle C L and the rectangle D E are together equal to the rectangle A L. Therefore the square on C B is equal to the rectangle A D, D B, together with the square on C D. THEOREMS FOR EXERCISE. 97. If a straight line, A B, be bisected in M and produced to O, the rectangle under the whole line, A O, thus produced, and the part produced, B O, together with the square on half the line, is equivalent to the square on the line M O, made up of the half and produced part. 98. If a straight line, A B, be bisected in M and divided unequally in O, the sum of the squares on the unequal parts A O and O B is equivalent to twice the square on half the line and twice the square on the intermediate part M O. 99. The square of the excess of one straight line over another is less than the squares of the two straight lines by twice their rectangle. 100. If two given straight lines be each of them divided into any two parts, prove that the rectangle contained by them is equivalent to the four rectangles contained by each part of the first and each part of the second. 101. If a straight line be divided into any two parts, the square on it is equivalent to the rectangles contained by the whole line and each part of the line. 102. If a straight line be divided into any three parts, the square on the whole is equivalent to the squares on each part, together with twice the rectangles contained. by the first and second, first and third, and second and third parts respectively. 103. If a straight line be divided into any two parts, the square on the whole line and one of the parts is equivalent to twice the rectangle contained by the whole and that part, together with the square on the other part. SECTION II.-The Theorem of Pythagoras. Enunciation. "If squares are described on the sides of a rightangled triangle, the two squares on the sides containing the right angle are together equal to the square on the side opposite the right angle." This theorem may be established in three ways. I. It may be shown that the two smaller squares can be cut into parts which may be made exactly to cover the larger square. II. It may be proved, as in Euclid's "Elements," by means of other theorems. III. It may be proved by means of the properties of similar figures. These will be discussed in a subsequent chapter; and it will be shown that the theorem is not only true for squares, but also for any other figures similar and similarly situated, as, for example, equilaterai triangles, semicircles, &c. I. Proof by Division and Superposition.-Let A B C be the triangle, and let squares be described on the sides. The smaller squares may be divided into parts which may be arranged in many ways so as to cover the larger square. (Figs. 137, 138, 139). Fig. 139 not only illustrates the theorem, but also shows that if a perpendicular be dropped from the right angle on to the hypothenuse, the square on a side is equal to the rectangle contained by the hypothenuse and that part of it between the side and the perpendicular. A simpler division of the larger square may be effected, as in Fig. 140. Let A HEh be the square on the side A h opposite the right angle of ▲ Ahd. Take hf on hd produced equal to Ad, and join ƒE. Then, by removing ▲ A d h to the position ADH, and AhfE to HFE, we obtain the squares on Ad and on C F, which is equal to dh. D FIG. 140. F E Algebraic Form of the Theorem.-The theorem thus proved may be stated algebraically as follows: "If a and b stand for the number of units in the sides AC, BC of a right-angled triangle and c for the number of units in the side opposite the right angle C, then Any two of these terms being given, the third can therefore be found by the ordinary rules. Example 1.-If the sides containing the right angle are respectively 65 and 72 inches, find the hypothenuse. Example 2.-If the hypothenuse and a side are respectively 109 and 91, find the other side. Theorems related to the above.—The theorem of Pythagoras is connected with two others, showing that "when the vertical angle is acute the square on the opposite side is greater than," and that "when the vertical angle is obtuse the square on the opposite side is less than the sum of the squares on the containing sides." To find how much greater or less it is we must project one of the sides containing the angle on to the other, and form the rectangle contained by one side and the projection of the other. Projection of one Line upon another. The projection of one line on another is the portion of the second which is intercepted by perpendiculars from the extremities of the first. A D FIG. 141. For instance, in Fig. 141, BD is the projection of C B on AB. Now the squares on A B and C B will be B together equal to the square on A C, plus the rectangle AB, BD when D falls on AB, and minus the rectangle A B, BD when D falls on the continuation of A B. By means of the Theorem of Pythagoras, a square may be found which shall be equal to the sum or difference of two given squares. When the sides of the two given squares are placed at right angles, the line joining their extremities will be the side of the square equal to their sum, and when the longer side is made the diameter of a circle and the other a chord meeting it, then the line joining their extremities will be the side of a square equal to the difference of the given squares. A square may always be found which shall be equal to the sum of any number of given squares. b d D C Let a, b, c, d be the sides of four squares (Fig. 142). Draw consecutively right-angled triangles, OAB, O BC, OCD, in which O A is equal to a, A B equal to b, BC equal to c, CD equal to d; OB will be side of a square equal to the sum of the first two; OC will be the side of a square equal to the sum of the first three; OD will be the side of a square equal to the sum of the four squares. B If all the squares are equal, the resulting square will be a multiple of one of the given squares; hence a square can always be found which shall be any required multiple of a given square. FIG. 142. To find a given Multiple of a given Square.—If the number representing the multiple be itself a square number, as, for instance, 16, then it will be sufficient to take a side a certain number of times (in this case four) the side of the given square, and construct a square upon it. If it be not itself a square number, it may always be broken up into two, three, or four square numbers, every number, as shown by Fermat's Theorem, being so divisible. Thus, 13 = 32 +22 747242 +32 8472 +52 + 32 + I. Consequently the multiplication of a square any number of times, can always be resolved into finding the sum of four, or less than four, |