the ends of the line are called the internal or external segments of the line, according as the point of section is in the line or the line produced. Prop. vi. Algebraically. Let AB contain 2a linear units, then its half BC contains a units; and let BD contain m units. Then AD contains 2a + m units, .. (2a + m) m + a2 = a2 + 2am + m3. But a+2am + n2 = (a + m)2, .. (2a + m) m + a2 = (a + m)2. That is, If a number be divided into two equal numbers, and another number be added to the whole and to one of the parts; the product of the whole number thus increased and the other number, together with the square of half the given number, is equal to the square of the number which is made up of half the given number increased. The algebraical results of Prop. v. and Prop. vI. are identical, as it is obvious that the difference of a + m and a m in Prop. v. is equal to the difference of 2a + m and m in Prop. vi, and one algebraical result expresses the truth of both propositions. This arises from the two ways in which the difference between two unequal lines may be represented geometrically, when they are in the same direction. In the diagram (fig. to Prop. v.), the difference DB of the two unequal lines AC and CD is exhibited by producing the less line CD, and making CB equal to AC the greater. Then the part produced DB is the difference between AC and CD, for AC is equal to CB, and taking CD from each, the difference of AC and CD is equal to the difference of CB and CD. In the diagram (fig. to Prop. vi.), the difference DB of the two unequal lines CD and CA is exhibited by cutting off from CD the greater, a part CB equal to CA the less. Prop. vII. Either of the two parts AC, CB of the line AB may be taken: and it is equally true, that the squares on AB and AC are equal to twice the rectangle AB, AC, together with the square on BC. Prop. vii. Algebraically. Let AB contain a linear units, and let the parts AC and CB contain m and n linear units respectively. That is, If a number be divided into any two parts, the squares of the whole number and of one of the parts are equal to twice the product of the whole number and that part, together with the square of the other part. Prop. VIII. As in Prop. VII. either part of the line may be taken, and it is also true in this Proposition, that four times the rectangle con tained by AB, AC together with the square on BC, is equal to the square on the straight line made up of AB and AC together. The truth of this proposition may be deduced from Euc. II. 4 and 7. For the square on AD (fig. Prop. 8.) is equal to the squares on AB, BD, and twice the rectangle AB, BD; (Euc. II. 4.) or the squares on AB, BC, and twice the rectangle AB, BC, because BC is equal to BD: and the squares on AB, BC are equal to twice the rectangle AB, BC with the square on AC: (Euc. II. 7.) therefore the square on AD is equal to four times the rectangle AB, BC together with the square on AC. Prop. VIII. Algebraically. Let the whole line AB contain a linear units of which the parts AC, CB contain m, n units respectively. Then m + n = a, and subtracting or taking n from each, ...m = a - n, squaring these equals, ... m2 = a2 - 2an + n*, and adding 4an to each of these equals, .. 4an + m2 = a2 + 2an + no. But a2 + 2an + n2 = (a + n)3, .. 4an + m2 = (a + n)®. That is, If a number be divided into any two parts, four times the product of the whole number and one of the parts, together with the square of the other part, is equal to the square of the number made of the whole and the part first taken. Prop. vIII. may be put under the following form: The square on the sum of two lines exceeds the square on their difference, by four times the rectangle contained by the lines. Prop. IX. The demonstration of this proposition may be deduced from Euc. II. 4 and 7. For (Euc. II. 4.) the square on AD is equal to the squares on AC, CD and twice the rectangle AC, CD; (fig. Prop. 9.) and adding the square en DB to each, therefore the squares on AD, DB are equal to the squares on AC, CD and twice the rectangle AC, CD together with the square on DB; or to the squares on BC, CD and twice the rectangle BC, CD with the square on DB, because BC is equal to AC. But the squares on BC, CD are equal to twice the rectangle BC, CD, with the square on DB. (Euc. II. 7.) Wherefore the squares on AD, DB are equal to twice the squares on BC and CD. Prop. ix. Algebraically. Let AB contain 2a linear units, its half AC or BC will contain a units; and let CD the line between the points of section contain m units. Also AD the greater of the two unequal parts contains a + m units, and DB the less contains a m units. · a2 + 2um + m2, and (a — m)2 = a2 2am + m2. That is, If a number be divided into two equal parts, and also into two unequal parts, the sum of the squares of the two unequal parts is equal to twice the square of half the number itself, and twice the square of half the difference of the unequal parts. The proof of Prop. x. may be deduced from Euc. 11. 4, 7, as Prop. ix. Prop. x. Algebraically. Let the line AB contain 2a linear units, of which its half AC or CB will contain a units; and let BD contain m units. Then the whole line and the part produced will contain 2a + m units, and half the line and the part produced will contain a + m units, .. (2a + m)' = 4a2 + 4am + m2, = 2a2 + 2(a + m)2. Hence .. (2a + m)2 + m3 That is, If a number be divided into two equal parts, and the whole number and one of the parts be increased by the addition of another number, the squares of the whole number thus increased, and of the number by which it is increased, are equal to double the squares of half the number, and of half the number increased. The algebraical results of Prop. Ix, and Prop. x, are identical, (the enunciations of the two Props. arising, as in Prop. v, and Prop. vi, from the two ways of exhibiting the difference between two lines); and both may be included under the following proposition: The square on the sum of two lines and the square on their difference, are together equal to double the sum of the squares on the two lines. Prop. XI. Two series of lines, one series decreasing and the other series increasing in magnitude, and each line divided in the same manner may be found by means of this proposition. (1) To find the decreasing series. In the fig. Euc. II. 11, AB and since AB. BH = AH3, .. (AH + BH), BH = A H3, .. BH2 = AH2 — AH. BH = AH. (AH – BH). If now in HA, HL be taken equal to BH, then HL2 = AH (AH - HL), or AH. AL = HL2 : that is, AH is divided in L, so that the rectangle contained by the whole line AH and one part, is equal to the square on the other part HL. By a similar process, HL may be so divided; and so on, by always taking from the greater part of the divided line, a part equal to the less. (2) To find the increasing series. From the fig. it is obvious that CF. FA = CA', Hence CF is divided in, in the same manner as AB is divided in H, by adding AF a line equal to the greater segment, to the given line CA or AB. And by successively adding to the last line thus divided, its greater segment, a series of lines increasing in magnitude may be found similarly divided to AB. It may also be shewn that the squares on the whole line and on the less segment are equal to three times the square on the greater segment. (Euc. XIII. 4.) To solve Prop. XI, algebraically, or to find the point H in AB such that the rectangle contained by the whole line AB and the part HB shall be equal to the square on the other part AH. Let AB contain a linear units, and AH one of the unknown parts contain x units, then the other part HB contains a − x units. And.. a (ax) = x2, by the problem, Whence x = 2 The former of these values of x determines the point H. It may be observed, that the parts AH and HB cannot be numerically expressed by any rational number. Approximation to their true values in terms of AB, may be made to any required degree of accuracy, by extending the extraction of the square root of 5 to any number of decimals. To ascertain the meaning of the other result = — In the equation a (a — x) = x2, for x writex, then a (a + x) = x3, √5+1 which when translated into words gives the following problem. • a. To find the length to which a given line must be produced so that the rectangle contained by the given line and the line made up of the given line and the part produced, may be equal to the square on the part produced. Or, the problem may also be expressed as follows: To find two lines having a given difference, such that the rectangle contained by the difference and one of them may be equal to the square on the other. It may here be remarked, that Prop. xI. Book 11, affords a simple Geometrical construction for a quadratic equation. Prop. XII. Algebraically. Assuming the truth of Euc. 1. 47. Let BC, CA, AB contain a, b, c linear units respectively, and let CD, DA, contain m, n units, then BD contains a + m units. And therefore, c2 = (a + m)2 + n3, from the right-angled triangle ABD, Prop. XIII. that is, c2 is greater than b2 + a2 by 2am. Case II. may be proved more simply as follows. therefore the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on CD; (II. 7.) add the square on AD to each of these equals; therefore the squares on CB, BD, DA are equal to twice the rectangle CB, BD, and the squares on CD and DA, but the squares on BD, DA are equal to the square on AB, (1. 47.) and the squares on CD, DA are equal to the square on AC, therefore the squares on CB, BA are equal to the square on AC, and twice the rectangle CB, BD. That is, &c. Prop. XIII. Algebraically. Let BC, CA, AB contain respectively a, b, c linear units, and let BD and AD also contain m and n units. Therefore c2 = n2 + m2 from the right-angled triangle ABD, and b n + (a — m)2 from ADC; ́¡¡ c2 — b2 = m2 — (a — m)3 Case II. that is, b2 is less than a3 + c2 by 2am. DC= = m — a units, 。。 c2 = m2 + n2 from the right-angled triangle ABD, and b2 = (m − a)3 + n2 from ACD, · b2 = m3 — (m − a)3, or b2 + 2am = a3 + c3, that is, b3 is less than a2 + c2 by 2am. Case III. Here m is equal to a. And b2 + a2 = c2, from the right-angled triangle ABC. • b2 + 2a2 = c2 + a3, that is, b is less than c* + a by 2a2, or 2aa. These two propositions, Euc. II. 12, 13, with Euc. 1. 47, exhibit the relations which subsist between the sides of an obtuse-angled, an acuteangled, and right-angled triangle respectively. |