CONS. 1 46. I. Pst. 1. On CD construct the square CEFD, join DE; through B draw BHG || CE, or DF; H " 5 Concl. DEM. 1 H. & 36. I. 3 Add. & Ax. 2. 4 Cons., Cor.4.II 5 Ax. 1. 6 Add. Cor. 4. II 8 Const. 9 Ax. 2. 10 Recap. then KLM || AD, or EF; and AK | CL, or DM; AM+ LG = square CF. CB, AL=CH; Adding to each ☐ CM, then ☐ AM = the But DM = DB; AM contained by and.. gnomon CMG=□AD.DB: But CMG and LG make up CEFD the square on CD; AD.DB+ CB2 = CD2. Wherefore, if a st. line be bisected and produced, &c. Q.E.D. Alg. & Arith. Hyp.-Let AC or CB=a=8; AB=2 a=16 & BD=m=4 AD=2a+m=16+4; and CD=a+m=8+4. Alg. Now AD=2a+m (× m) and therefore (2a+m) m=2am+m2. (Add a2) therefore (2a+m) m+a2=a2 +2 am+m2 But a2+2 am+m2=(a+m) 2 therefore (2a+m) m+a2=(a+m)2 Arith. Now AD=16+4 (x4) then (16+4) × 4=(2 × 8 × 4) + (4 × 4)=64+ 16. (Add 8 × 8) and 4 (16+4)+64 64+64 +16 But 64+64 +16=(8+4)2=144 therefore 4 (16+4)+64=(8+4)2=144 COR.-If a line AD be drawn from the vertex A of an isosceles triangle to the base or its production, the difference between the squares of this line and the side of the triangle is the rectangle under the segments BD DC of the base. For (by Cor. 4, P. 47. I.) A D2 ~AC2 = CE2~DE2; but (by 5 and 6. II.) CE2~DE2BD.DC, . AD2~AC2 = BD. DC. N.B. If AD coincide with the perpendicular A E, the part DE will vanish. SCH.-The algebraical results of Prop. 5 and 6 differ only in appearance, "arising from the two ways in which the difference between two unequal lines may be represented geometrically when they are in the same direction." In Prop. 5, the difference DB-a-m-10-6, of the two lines AC=a=10 and CD=m=6, is exhibited by producing the less CD, and making CB=CA. Then DB AC or CB-CD-a―m—10—6—4. In Prop. 6, 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; then DB =CD-CB or CA=(a+m)—a—(8+4)—8—12—8—4. USE AND APP.-MAUROLICO, a mathematician, and abbot of Messina, who died A.D. 1575, measured the diameter of the earth by making use of this Proposition. From A the top of a mountain, the height of which AD is known, the angle CAB is measured, formed by EA and AB, a tangent to the circle in B, the limit of vision. In the rt. angled triangle ADF, the angles being obtained, the sides AF, FD will be found by Trigonometry; and FD equals FB: thus AB=AF+FB, and its square=AB2. Now (by Prop. 6. II.) ED being divided in C and produced to A, the rectangle EA.AD+CD2 =AC2; and (by Prop. 18. IIÏ.) ABC is a right angle therefore AC2=AB2 + BC2 or CD2, and thus the rectangle EA.AD+CD2=AB2+CD2; : B A F D E C take away the common part CD2, and the rectangle EA AD AB2. Divide both sides of the equation by AD, and EA= the diameter of the earth. AB2 Then EA-AD-ED Example. A mountain is 21 miles, AD, above the level, FD, of the sea; and the limit of vision, or AB, is 141 miles: required the earth's diameter DE. By the result just obtained EA= and ED EA-AD. AB2 7949.9 the earth's diameter. PROP. 7.-THEOR. If a st. line be divided into any two parts, the squares of the whole line and of one of the parts are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. CONS. 46. I., Pst. 1 and 31. I. DEM. 43. I., EXP. 1 Hyp. Axs. 2 and 6, Def. 30. I., Cor. 4. II., Ax. 8, and 34. I. Let AB be divided A AB and CB = twice AB.BC+AC2. CONS. 146. I. Pst. 1. On AB make a square 4 Concl. DEM. 1 43. I. 2 Add. & Ax.1 3 Ax. 6. 5 Ax. 1. " HK AB; then the squares AE + CK = twice AK + HF. the compl. AG = the compl. GE; But AKCE make up AKF+CK; 8 Ax. 1 & 8... the gnom. AKF + CK=twice the rect. AB.BC: 9 Add. & 34. I Adding to both the equals HF, i. e., the square on HG or AC, 10 Ax. 2. 11 Const. 12 C. 1, 3. 13 Concl. 14 Recap. then AKF+CK+HF= twice AB. BC and But AKF+CK+ HF make up the figures and ADEB and C K are the squares on AB .. AB2 + BC2 = twice AB. BC+AC2. Wherefore, if a st. line be divided into any two parts, &c. Q.E.D. Alg. & Arith. Hyp.-Let A B a linear units=16; AC=m=9 and CB =n=7. Alg. Then a m+n; Squaring, a2=m2 +2mn+n2. (Add n2) a2 + n2=m2 + 2 m n+2 n2. But 2 mn+2 n2=2 (m+n) n=2 an, therefore a2+n2=2 an+m2. Arith. 16=9+7; Squaring, 256=81+126+49. (Add 49) 256+49=81+126+98. But 126+98=2 (9+7) 7=2 (16 × 7)=224, therefore 256+49=222+81=305. Another form of stating the same result is, Let A B=a=16; AC=b=9; and BC-a-b—16—9—7. then A B2 = a2 BC2a2-2ab+b2 Sum 2 a2-2ab+b2 AC2= b2 2 a2-2ab+b2 COR. 1.-If AB and BC be considered as two independent lines, AC being their difference, "the sum of the squares of any two lines is equal to twice the rectangle under them together with the square of the difference;" i. e., AB2 + B C2 = 2 AB. BC + AC2; or 100+ 64 = (2 × 80) + 4. COR. 2.-Hence and (by 4. II.) the square of the sum of two lines, the sum of their squares, and the square of their difference, are in arithmetical progression,-the common difference being twice the rectangle under the sum. By 4. II. (AB+BC)2=AB2+BC2+2 AB. BC; and by Cor. 7. II. A B2+BC2=(AB-BC)2+2 AB. BC; or AC2+2 AB. BC. Thus the common difference is 2 AB.BC; therefore the quantities (AB+BC)2, (AB2+BC2), and A C2 are in arithmetical progression; as 324, 164, and 4,—the com. dif. being 160. PROP. 8.-THEOR. If a st. line be divided into any two parts, four times the rectangle contained by the whole line and one of the parts together with the square of the other part, is equal to the square of the st. line which is made up of the whole and that part. M CONS.-Pst. 2. A terminated st. line may be produced in a st. line. 3. I. From the greater line to cut off a part equal to the less. 46. I. On a given st. line to describe a square. 31. I. Through a given point to draw a st. line parallel to a given st. line. DEM.-34. I. The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them. Ax. 1. Magnitudes equal, &c. 36. I. Parallelograms upon equal bases and between the same parallels are equal. 43. I. The complements of the parellelogram which are about the diameter of any parallelogram, are equal to one another. 4. II. If a st. line be divided into any two parts, the square of the whole line equals the square of the two parts together with twice the rectangle contained by the parts. Def. 30. I. A square is a four-sided figure having all its sides equal, and its angles rt. angles. Cor. 4. II. The parallelograms about the diameter of a square are also squares. = BC; HLF 246. I. Pst. 1 On AD describe the square A EFD, and join 3 31. I. 5 Concl. ED; through B and C draw ||s BL, CH, to AE or DF, and cutting ED in the points K and P; also through K and P, MGKN and XPRO Is to AD or EF; then AK+MR+ (HR+BN) +NL+XH, fill up the figure AEFD. |