PROP. VII. A B2+ B D2 = 2 AB.BD+ AD3, or 100+9 = (2 × 10 × 3)+49= 109. Cor. 1. AB+BD2 = 2 AB.BD+(AB-BD); or 100+9= 60 +(10—3)2 = 109. 2. (AB+BD); (AB+ BD); and (AB-BD) are in Arithmetical progression, the common difference being 2 AB.BD; or, 169, 109, and 49; the com. dif. being 2 × 10 × 3 = 60. = PROP. VIII. 4 AB.BD+AD (AB+BD); or 4 (10 x 3)+49= (10 × 3)2 = 169. Cor. 1. (AD+DB) = 4 AD.DB+(AD-DB)2; or (7+3)3 = 4 (7x3)+(7-3)=100. 2. 4 (AD+DB)2 = 4 AD.DB+4 (ADD); or 4 × 25 2 (4 × 21) + (4 × 4) = 84+16. CASE III.-Let a line be divided equally and unequally, AB 10 representing the whole line; AC or CB= 5, the half line; and AD=7; DB=3, the unequal parts. PROP. V. AD.DB+CD3 = CB3; or (7 × 3) + (2 × 2) = 5 × 5; i.e., 21+4=25. N.B. The arith. mean = AD+DB 2 2 =AC; the com. dif. AD-DB · (AD+DB)2 = AD.DB+(ADD); or 25=21+4. Or, Let AD and DB be regarded as two independent lines; then AD.DB+(ADDB) = (AD+DB)2; or (7 × 3) + 4 = (1+3)3 = 25. 2 Again, (AD+ DB). (AD-DB) + D B2 = A D2; or (10 × 4)+9= 7 x7=49. 3 or, AD-DB= (AD+DB). (AD-DB; or 49-9=10 × 4 = 40. x Cor. AC-CD (AC+ CD) (AC-CD); or 25-4=7 × 3 = 21. LARDNER'S Cor. 1. The rectangle is a maximum when AB is bisected by D; its maximum value = (4B)2 2 2. The sum of the squares of the parts is a minimum when AB is bisected: the minimum value being 2 (4B)* 2 3. Of all rectangles having the same perimeter, the square contains the greatest area. 4. Of all rectangles equal in area, the square is contained by the least perimeter. 5. If a perpendicular be drawn from the vertex of a triangle, as in Prop. 13, to the base (AB+ AC). (AB ~AC) = (BD+DC). (DC~BD), or (7.8102 + 6.4031) (7.8102-6.4031)=(6+4) x (6-4)=20. 6. The difference between the squares of the sides of a triangle, as in Prop. 13, is equal to twice the rectangle under the base and the distance of the perpendicular from the middle point, x, of BC: i. e., A B2-AC2 = 2 BC.Dx; or 61-41= 2 × 10 × 1 = 20; or, 61—41 = (2 × 2) × 5 = 20. N. B.-If the perpendicular AD falls within the base, PROP. IX. AD2+ DB2 = 2 (A C2+ CD); or 49 +9=2 (25+4)= 58. ᎪᎠ 2 Or A D2 + DB2 = 2 (AD+DB)2+2 (AD-DB); or (2 × 25) + CASE IV.-Let a line be bisected and produced, AB = 10 representing the original line; AC or CB= 5, its half; and BD = 3 the part produced. PROP. VI. AD.DB + C B2 = C D2; or (13 × 3) + (5 × 5) = 64. Cor. Let A B = 10, the base of an isosceles triangle, be bisected in C, and produced to D; join ED = 11.7898; then E D'-E B2 = AD.DB; or 139-100 = 13 × 3 = 39. A C BD PROP. X. AD2+ DB2 = 2 (A C2 + CD2); or 169+9=2 × (25 +64) = 2 × 89 = 178. CASE V.-A line is cut in extreme and mean ratio, when the rectangle of the whole line and one segment equals the square of the other segment. بعا PROP. XI. AB so cut in H that AB.HB = AH2 F 2. When a line CF, or its equal, is cut in E G H C K D extreme and mean ratio, the rectangle AC. (A C— AF) = AH3 or AF2; or AC.HB = Å H2. 3. A line CF being cut in extreme and mean ratio, AC.AF AC-AH2; i. e., 10 × 6.18034= 100— 38.1966 61.8034. CASE VI.-The measure of the square of the side of an obtuseangled triangle opposite to the obtuse angle. PROP. XII. A B2=B C2+ AC2+2 BC. CD, 100 = 25+ 45+ (2 × 15). or, A B2 > BC2+ AC by 2 BC.CD, 100 > 25+ 45 by 30. N.B. The Area of a triangle may be ascertained when the three sides are known; for Area= AD.BC A B C D CASE VII.-The measure of the square of the side subtending an acute angle. AM G DCBC DB C PROP. XIII. 1o CB2+AB2 = AC2+2 BC. BD; 100+61=41 +120. or AC (CB2+ AB) by 2 BC.BD; 41 < 161 by 120. 2o. A B+ BC2=AC+2 BD.BC; 61+4=41 + (2 × 12) ; or A C2 < (A B2 + BC2) by 2 BD.BC; 41 < 65 by 24. 3o. A C2 < (A B2 + BC2) by 2 BC.BC; 25< 41+ 16 by 2 × 16. If in 2o a perpendicular CG be drawn from C to AB, the □ AB. GB=BC.DB; or 7.8012 × 1.5364=12=2 × 6. CASE VIII-A square is found equal to a given rectilineal figure. PROP. XIV. By Const., BCDE=A; BE.EF+EG2 = G F2 = GH2; 9×4+6.25 = 42.25; BD=EH2 = A. /366 EH side of the square. PRACTICAL RESULTS. The Practical Results at which we arrive from studying the principles demonstrated in the first and second Books of Euclid, are among the most useful and important of any belonging to Geometry when united to a few leading theorems of the third and sixth books, they constitute by far the most fruitful source of scientific progress; they are, in fact, the foundation on which our after reasonings depend respecting all magnitudes, space, and number, the form and motions of the heavenly bodies,and all the laws by which the material universe is governed. Nearly all the Problems in Euclid's Six Books of Plane Geometry may be derived directly and immediately from the first and second books. Indeed, for the simple construction of Geometrical Figures, scarcely any principles are demanded which are not to be found in, or naturally inferred from, those books. More fully therefore to give the Learner an idea of the wide application of the knowledge he may have acquired, I will briefly exhibit the Practical Results to which I believe we have attained: they may be arranged under four leading divisions :— 1st. The Problems for the Construction of Geometrical Figures both stated and demonstrated in Books I. and II.; 2nd. Problems in the other books which, though demonstrated under their respective Propositions, are yet most intimately connected with the first and second, for their construction and solution; 3rd. Principles of Construction for Geometrical Instruments to measure lines and angles; and for Geometrical figures to exhibit the representative values of actual magnitude and space; and 4th. Principles which without requiring that we should measure all the boundaries of a Surface, yet enable us accurately to calculate distances, magnitudes, and areas. |