Alg. (47. I.) b2 + a2=c2 Add a2 and b2 + 2 a2=c2 + a2 Arith. (47. I.) 25+16=41 Add 16 and 25+32=41 +16 COR. I.-If in the figure to CASE II. a perpendicular CG be drawn from the angle C to AB, the rectangle under the side AB and the part GB intercepted between the perpendicular and B is equal to the rectangle of BC.DB. The two rectangles AB.BG and BC.DB each equal half the difference between the (AB2 + BC2)-AC2 square AC and the squares AB and BC; i. e., 2 SCH.-The Propositions 12 and 13 are of high importance, for they contain the elements of Trigonometrical Analysis, or the Arithmetic of Sines. USE AND APP.-1. In finding the Area of a Triangle, it is of the greatest advantage to obtain the perpendicular either by calculation or by measurement. When the three sides of a triangle are given, the perpendicular may be obtained, on the foregoing principles, in either of the ways following: FIRST METHOD. When the perpendicular, AD, falls within the base. By 12. I., from A draw AD perpen. to BC; 47. I., in triangle ABD, AB2=AD2+BD2 47. I., & in triang. ACD, AC2=AD2 + CD2 Subtract, and AB2-AC2BD2-CD2 By 3 & 10. I. From BC cut off BE equal to DC, and bisect ED in G; ED is evidently bisected in G, so that 2 GD =ED. B A EGD Now, since, by Cor. 5. II., the difference of the squares of two lines equals the rectangle under their sum and difference; AB2-AC2 (BA+AC) (BA-AC)=(BD+DC) (BD-DC)=BC × 2 GD=2BC × GD. Dividing now both sides of the equation by 2 BC, we have Ex. Given BC=5, AB=4, and AC=3; required the perpendicular AD. SECOND METHOD. — When the perpendicular, AD, falls without the base. By 12. I., from ▲ A draw AD perpendicular to BC produced; 47. I., in triangle ABD, AB2=AD2 + BD2 47. I., and in triangle ACD, AC2=AD2 +CD2 Subtract, and A B2-AC2=BD2-CD2 By 10. I., let BC be bisected in G; then BD+CD= B A Divide by 2 BC,—and GD (AB+AC) (AB—AC) = 2 BC Now CD=GD-GC; And (47. I.) AD=√AC2—(GD—-GC)2 Ex. Given the three sides BA=5, AC=4, and BC=2; required the perp. AD. 2. When the intercepts between the foot of the perpendicular and the angles on the base are ascertained, the Area of the Triangle is by the First BD+DC Method 2 × AD; and by the Second Method BD-DC × AD; that is, half the rectangle of the base multiplied by the perpendicular. PROP. 14.-PROB. To describe a square that shall be equal to a given rectilineal figure. SOL. 45. I. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given angle. Def. 30. A square, &c. Psts. 2 and 3; 3. I. 10. I. To bisect a given finite st. line. Pst. 1. DEM.-5. II. 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 equals the square of half the line. Def. 15 and 16. A circle is a plane figure contained by one line called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another. And this point is called the centre of the circle. 47. I. The square of the hypotenuse, &c. Ax. 3. If equals be taken, &c. Ax. 1. Things equal, &c. EXP. 1 Datum. 2 Quæs. Let A be the given rectil. to make a square equal to the fig. A. A H G CONS. 1 45. I. Describe a right B C BCDE=A: D 2 Def. 30. 3 Sup. 4 Pst. 2 & 3. I. 5 10. I. Pst. 3. If the side BEED the fig. is a square; but if BEED, produce BE to F, and make EF =ED: 6 Pst.2 & 1. produce DE to H, and join GH: DEM. 1 C. 5. 2 5. II. Def.16. 4 Sub. & Ax. 5 C.1. Ax. 1. 6 Concl. 3 BF is bisected in G, and cut unequally in E; BE. EF + EG2 = GF2, and G F2 = GH2 : But the square on GH = EH2 + EG2; taking away EG2, then BE. EF=EH2: But BD=BE.ED, or BE. EF; therefore BD = the square on EH; 7 C.1 & D. 6. But BD=the fig. A: :. EH2 = A. Wherefore, a square has been made equal, &c. 8 Recap. Q.E.F. Alg. & Arith. Hyp.—Given the fig. A=ab=36, in area; required the side EH=x of a square equal in area. Let the lines, EH=x; BE one side of the rectangle=6=9, and ED the other side=a=4. Alg. By the problem, ab=x2 (therefore, (√) √ab=x the side of the square. Arith. 4x9= x2 and 36x=6 the side of the square. USE AND APP.-1. By this proposition we may find a mean proportional to two given lines, as demonstrated in Prop. 13, bk. vi. Given the lines BE and EF; their sum=2r=13; BE=x=9; EF= 2r-x=13-9=4: required to find y, a mean proportional to x, and 2r--x; i. e., so that xy::y:2r-x. Bisect BE in G; GF=r=6.5 By 5. II. x.(2 r—x) + (x—r)2 =p2 But GF2=r2=y2 + (x—r)2 Subtract (x-r)2 And x. (2r-x) =y2 (9×4(+6.25-6.52=42.25 42.25-y2+6.25 9x4=36=y2 Extract the √; y=√36=6 the mean proportional between 9 and 4. Briefly, Multiply the two numbers, and take the square root of their product for the mean proportional. 2. The same Prop. 14 serves also for approximating to the square of curve-lined figures, and even of the circle itself; for the circle may be regarded (as in 41. I., Use 4) as a polygon with an infinite number of sides ;-and if we reduce this polygon to a square, we nearly obtain the square of the circle. 3. The Areas of all plane figures are calculated in squares; and by means of reducing any right figure, first into a rectangle, and then into a square, we obtain the ready means of finding the Area in the units of surface. N.B. A selection of the most useful and remarkable problems and theorems, which may be inferred from the principles developed in the second book, will be found in the APPENDIX to the Gradations in Euclid, SERIES II., Problems and Theorems, bk. ii. REMARKS ON BOOK II. 1. Of the fourteen Propositions of this book, the ten first contain the theory of the relations of the rectangles and squares on divided lines; the twelfth and thirteenth the theory of the relation between the square of any one side of a triangle, and the squares of the other two sides, whatever be the angles,-and constitute the foundation of the Arithmetic of Sines. 2. Lines cut into any two parts are considered in Propositions 2, 3, 4, 7, and 8. 3. Lines cut into two equal and two unequal parts,-in Propositions 5, 6, 9, and 10. SYNOPSIS OF BOOK II. CASE I.-Given two lines, as A undivided = 10, and BC = 60 divided into the parts BD=30, DE = 20, and EC = 10. PROP.I. A.BC=A.BD+A.DE+A. EC; or 10 × 60 = (10 × 30) + (10 × 20) + (10 × 10) = 600. Cor. 2 A. BC; or 3 A. BC; or 4 A. BC, &c. = A.BC. 10 × 60 600. CASE II.-Let a line be divided into any two parts; AB = 10, representing the whole line; AD=7, and DB=3, the unequal segments. PROP. II. AB. AD+AB.DB= AB2; or (10 × 7) + (10 × 3) = 10 × 10 = 100. PROP. III. AB.BD (3 x 3)=30. = AD.DB + DB2; or (10 × 3) = (7 × 3) + Cor. 1. AB2 DB2 = (AB+DB) (AB-DB); or 100-9= 13 × 7=91. 2. AD2-DB2> (AD-DB)2 by 2 DB. (AD-DB); or 49-9 > 16 by 2 × 3 × 4 = 24. PROP. IV. AB2 = AD2 + D B2 +2 AD.DB; or 100 (2 × 7 × 3) = 49 + 9 + 42. = 49+9+ Cor. 1. Parallelograms about the diam. of a square are also 4. When a line is divided into any number of parts, as x=5; y = 3; z= 2; then A B2 = x2 + y2 + z2 + 2 (x.y + x.z+y.z) or, 100=25+ 9 + 4 + 2 (15 + 10 + 6) |