reason are Deu 10. Concl. .. also CGKB is rec- A с B tangular : 11 D. 6 & 10. :. fig. CGK Bis a square and on CB. H K 12 Sim. 34. I. For the same HF is a square on ; D E k or AC.CB; AC.CB: 18 D. 13. And figures HF, CK are squares on AC and CB; 19 D.17&18., figures HF + CK + AG + GE = AC? + BCP + 2 O AC.CB. 20 Const. But HF, CK, AG, and GE make up ADEB, the square on AB; 21 | Ax. 1. :: A B2 AC + BC2 + 2 AC.CB. 22 Recap. Wherefore, if a st. line be divided into any two parts, &c. Q.E.D. Alg. & Arith. Hyp.--Let AB = a = 12; AC = m = 8; and BC = n = = 4. Alg. Then a = m +n Arith. Then 12 = 8 + 4 Squaring, a? (m + n)?, Squaring, 122 (8 + 4)2 = 144, or m2 + 2 mn + na or 64 + (2 x 32) + 16 Or, Let the line be made up of the parts k + y. Then (2. II.) A? = A.k + A.y. But (3. II.) A. k = ko + k.y: And A .y = y2 + k.y. Hence A? ka + y2 + 2 k. y. Cor. 1.—The parallelograms about the diameter of a square are х also squares. 2.—The square of a st. line is four times the square of its half; for AC being equal to CB, AC + CB2 = 2 AC? or 2 CB?; and the rectangle AC.CB is the same as AC? or CB2. Thus AB = 4 (41)", i.e., in the numerical example 12 x 12 = 4 (6 x 6) = 144. Х n + 1 3.—Half the square of a st. line is equal to double the square of half 12 x 12 the line; thus (43) = 2 (45) or = 2 (6 x 6); or 1 = 2 x 36 = 72. 2 4.- This proposition may also be employed for a st. line divided into any number of parts; then, the square of the st. line will be equal to the sum of the squares of all the parts, together with double the rectangle under every distinct pair of parts. Let AB = k + y + 2; then A B ka + y + 32 + 2 (k.y + y .2 + k.2). Take a number 12 = 6 + 4 + 2; then 12 x 12 =(6 x 6) + (4 x 4) +(2 2) + 2 {(6 x 4) + (4 × 2) + (6 x 2)}; or 144 = 36 + 16 + 4+2 (24 + 8 + 12). USE AND APP.-1. In Algebra, the square of a binomial, that is, of a quantity made up of two terms, as k+y, is identical with the squares of the parts added to twice the product of the parts ; thus (k + y)= ka + y2 + 2 k. y. 2. This Proposition points out a practical way of extracting the Square root of a number. Let a number 144 be represented by the square A E, and its square root by the st. line M A CB AB. The square root of every number G containing an even number of digits, con K sists of half that even number of digits ; L H and the square root of every number containing an odd number of digits, consists of digits. We know therefore that AB 2 the line required will be represented by a D F E number consisting of two digits. Suppose AB divided at C, so that AC represents the first digit, or digit of higher value, and C B the second, or digit of lower value. Seek the root of the value of the first figure in 144, namely, of 100,--and it is found to be 10 ; the square of 10 therefore is represented by HF; HF being removed or subtracted, there remains 44 for the sum of the rectangles, AG, FK and of the square CK. But as the gnomon AKF is not convenient for use, the rectangle KF may be removed and placed as a continuation of AG, as MLHA; the whole rectangle LB therefore contains the remainder 44. As AC equals 10, twice AC, or MC= 20. We must therefore divide 44 by 20; that is, to find the divisor we double the Root already found, namely 10, and say,--how many twenties are there in 44 ! We find there are two for the side BK; but since 20 was not the whole side MB, but only a part MC, the new figure in the quotient, 2. must be added to the divisor 20, making 20 + 2, or 22 ; and 22 is contained exactly in 44. The first digit in the root is 1 ; the second, 2 ; or together, 12. Thus the Square of 144 is equal to the square of 10, to the square of 2, and to twice 20, which are the two rectangles comprehended under 2 and 10. It is for these reasons that the formula or rule given for extra the square root of a number, requires the number to be separated into periods of two; the nearest square, and its root, to the left hand period are found,-and as twice the rectangle of the parts added to the square of the part to be found, make up the remainder,-for the division of that remainder we take twice the root, or rectangle already found, and to complete the divisor add to that twice the root the new term of the root : and so on, until the operation is completed. PROP. 5.-THEOR. If a st. line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts together with the square of the line between the points of section, is equal to the square of half the line. CON.–46. I., Pst. 1, and 31. I. of any parallelogram, are equal to one another. are equal to one another. squares. another, and the diameter bisects them. Exp. 1 Hyp. I. 2. 2 Concl. Let st. line AB be divided into two equal parts in. C, and into two unequal parts in. D; then the AD.DB + the square on CD the square on CB. Cons. 1 by46.1.Pst. 1 On CB construct a square CEFB, and join BE; through H draw | KLM || CB or EF; and through A draw | AK || CL or BM; 5 Concl. then O AH + square on CD = square EB, i.e., the on CB. O OS DEM. 1 | by 43. I. The complem. CH = the complem. HF; 2 Add. &Ax. 2. adding ODM to each, then a CM = DF: But :: AC = CB; then D AH = o SDF and CH. and AH is contained by AD, DH; 7 Concl. :, AH also AD.DB: 8 Def. 2. II. Also os DF and CH make the gnomon CMG; 9 | Ax. 1. :: gnomon CMG the CAD.DB: 10 Add Cor.4.11 Add the square LG, that is, the square on CD; 11 Ax. 2. then gnomon CMG + LG O AD.DB + CD: 12 Cons. But CMG + LG make up CEF B, or the square on CB; 13 As. 1. ..O AD.DB + C D? CB? 14 Recap. Wherefore, if a st. line be divided into two equal Q.E.D. parts, &c. ma 12 Aritk. & Alg. Hyp. -Let AC = CB = a = : 10, and AB=AC + C B = 2a = 20, or AB=AD+ D B, and let C D=m=6. Then AD = a + m = 10 + 6, and DB = a-m=10 - 6. Alg. Now m= (a + m) - (a – m) 2 and (a + m) (a – m) (+ m2) :: (a + m) (a – m) + m2 = a* Arith. 6 = (10 + 6) - (10-6) 2 (+ 36).. (16 x 4) + 36 = 100 COR.—It is manifest that the difference of the squares of two unequal st. lines, AC, CD, is equal to the rectangle contained by their sum and difference ; i.e., AC - CD = (AC + CD) (AC - CD). From the square EB on CB take the square D M on DB, there will remain os EM and CH; if HF be taken from EM and placed by the side of CH, as AL, then CH and HF become AH, and D A H is contained by AC + CD or AD, and by AC – CD or DB: hence AC – CD, i.e., the difference of the squares, = (AC + CD) (AC - CD), i.e., the rectangle contained by their sum and difference. Since AC = a = 10 is half the sum of AD + DB, and CD= m = 6 half their difference, the corollary inay be thus expressed :-" The rectangle contained by two st. lines A D and DB, is equal to the difference of the squares of half their sum AC, and half their difference CD;" i.e., AD. DB = AC - CD?, or (a + m) (a – m) = a? - m?. 2 Lardner's Corollaries to this Proposition are e-1st. As the intermediate part CD diminishes, the rectangle increases, and vice versa. The rectangle is a maximum when AB is bisected by D, its value being (AB)".—2nd. The greater the rectangle is, the less will be the sum of the squares of the parts; the sum of the squares of the parts being at a minimum when the line CB is bisected in D.3rd. Of all rectangles having the same perimeter, the square contains the greatest area.—4th. Of all rectangles equal in area, the square is contained by the least perimeter.—5th. If a perpendicular be drawn from the vertex of a triangle to the base, the rectangle under the sum and difference of the sides is equal to the rectangle under the sum and difference of the segments.—6th. The difference between the squares of the sides of a triangle, is equal to twice the rectangle under the base and the distance of the perpendicular from the middle point. These Corollaries may be taken as the subjects of Geometrical Exercises. Sch.— When a st. line is divided in two points equally and unequally, several linear magnitudes have to be considered ; Ist, the whole line AB; 2nd, the equal segments AC and CB; 3rd, the unequal segments AD and DB; 4th, the intermediate part, that is, the part CD between the points of section. The following are the principal properties connected with the equal and unequal division of a st. line 1°. The intermediate part equals half the difference of the unequal parts; thus, CD=} (AD - DB), or 6 = 1 (16 – 4). 2°. The greater segment equals half the sum added to half the difference; thus, AD = } (AD + DB) + } (AD - DB); or 16 = }(20 + 12). 3o. The less segment equals half the sum minus half the difference ; thus DB=} (AD + DB - 1 (AD- DB), or 4= 1 (20 – 12). 4°. The sum of two unequal lines equals twice the less added to the difference of the lines ; thus, AD + DB= 2 DB + (AD – DB); or 16 + 4 = (2 x 4) + (16 – 4). = USE AND APP.—1. By the Cor. to this Proposition the difference between the squares of two unequal numbers may be found without squaring them ; for the product of their sum and difference will equat the difference of the squares. Thus, if in a rt. angled triangle the hypoten use is 100 feet, and the base 80, we may find the perpendicular without squaring the numbers ; for (100 + 80) x (100 – 80)=3600, and V3600 = 60 = the perpendicular. i 2. We have also the means of finding Quantities in Arithmetical Progression. To be in Arithmetical Progression, quantities must increase or diminish by a |