meter AB; the circumference is greater than three times AB, 10 of AB, but greater than of AB. 71 D In the circle ABD apply the straight line BD equal to the radius BC. Draw DF perpendicular to BC, and let it meet the circumference again in E; draw also CG perpendicular to BD; produce BC to A, bisect AC in H, and join CD. A B H C F E It is evident that the arcs BD, BE are each of them one-sixth of the circumference (IV. 15 Cor.), and that therefore the arc DBE is one-third of the circumference. Wherefore the line (I. Sup. 6) CG is a mean proportional between AH, half the radius, and the line AF. Now, because the sides BD, DC of the triangle BDC are equal, the angles DCF, DBF are also equal; and the angles DFC, DFB being equal, and the side DF common to the triangles DBF, DCF, the base BF is equal to the base CF, and BC is bisected in F. Therefore, if AC or BC=1000, AH=500, CF=500, AF=1500, and CG being a mean proportional between AH and AF, CG2= (VI. 17) AH.AF=500×1500-750,000; wherefore CG-866.0254+. Hence, also, AC÷CG=1866.0254+. Now, as CG is the perpendicular drawn from the centre C, on the chord of one-sixth of the circumference; if P=the perpendicular from C on the chord of one-twelfth of the circumference, P will be the mean proportional between AH (I. Sup. 6) and AC+ CG, and P2=AH (AC+ CG) = 500 × 1866.0254 +=933,012.7+. Therefore, P=965.9258+. Again, if Q=the perpendicular drawn from C on the chord of one twenty-fourth of the circumference; Q2=AH (AC+P)= 500×1965.9258+=982,962.9+; and, therefore, Q=991.4449+. In like manner, if S be the perpendicular from C on the chord of one forty-eighth of the circumference, S2=AH (AC+Q)=500× 1991.4449+=995,722.45+ ; and S=997.8589+. Lastly, If T be the perpendicular from C on the chord of one ninety-sixth of the circumference, T2 = AH (AC+ S) = 500x 1997.8589+, =998,929.45+, and T=999.46458+. But, by the last proposition, the chord of one ninety-sixth part of the circumference is a mean proportional between the diameter and the excess of the radius above S, the perpendicular from the centre on the chord of one forty-eighth of the circumference. Therefore, the square of the chord of one ninety-sixth of the circumference AB (AC-S)=2000×2.1411-, =4282.2-; and therefore the chord itself 65.4386-. Now, the chord of one ninetysixth of the circumference, or the side of an equilateral polygon of ninety-six sides inscribed in the circle, being 65.4386–, the perimeter of that polygon will be = (65.4386–) 96=6282.1056-. Let the perimeter of the circumscribed polygon of the same number of sides be M, then (I. Sup. 2, Cor. 2) T: AC:: 6282.1056– · M, that is (since T = 999.46458 +, as already shown), 999.46458+: 1000 :: 6282.1056-: M. But 999.46458: 1000:: 6282.1056: 6285.461-; therefore the perimeter of the polygon circumscribed about the circle is less than 6285.461; now, the circumference of the circle is less than the perimeter of this polygon; wherefore it is less than 6285.461 of those parts of which the radius contains 1000. The circumference, therefore, has to the diameter a less ratio (V. 8) than 6285.461 has to 2000, or than 3142.7305 has to 1000; but the ratio of 22 to 7 is greater than the ratio of 3142.7305 to 1000, therefore the circumference has a less ratio to the diameter than 22 has to 7, or the circumference is less than 22 of the parts of which the diameter contains 7. It remains to demonstrate, that the part by which the cir10 cumference exceeds three times the diameter is greater than 71 of the diameter. It was before shown, that CG2=750,000; wherefore CG= 866.02545-; and AC+CG=1866.02545-. Also, P2-AH (AC+CG) = 500×1866.02545-933,012.73—; and P-965.92585-. -= Again, Q2=AH (AC+P) =500×1965.92585 — — 982,962.93— ; and Q=991.44495—. In like manner, S2 = AH (AC + Q) = 500 × 1991.44495 995,722.475-, and S=997.85895-. But the square of the chord of the ninety-sixth part of the circumference AB (AC-S)=2000 (2.14105+)=4282.1+, and the chord itself 65.4377+. Now, the chord of one ninety-sixth part of the circumference being =65.4377+, the perimeter of a polygon of ninety-six sides inscribed in the circle = (65.4377+) 96= 6282.019+. But the circumference of the circle is greater than the perimeter of the inscribed polygon; therefore the circumference is greater than 6282.019 of those parts of which the radius contains 1000; or than 3141.009 of the parts of which the radius contains 500, or the diameter contains 1000. Now, 3141.009 has to 1000 a greater ratio than 3+ to 1; therefore the circumfer 10 71 10 ence of the circle has a greater ratio to the diameter than 3 +71 has to 1; that is, the excess of the circumference above three times the diameter is greater than ten of those parts of which the diameter contains 71; and it has already been shown to be less than ten of those of which the diameter contains 70. Therefore, &c. Q. E. D. COR. 1. Hence the diameter of a circle being given, the circumference may be found nearly, by making as 7 to 22, so the given diameter to the circumference. COR. 2. As 7 to 22, so is the square of the radius to the area of the circle nearly. For it has been shown, that (I. Sup. 4, Cor) the diameter of a circle is to its circumference as the square of the radius to the area of the circle; but the diameter is to the circumference nearly as 7 to 22, therefore the square of the radius is to the area of the circle nearly in that same ratio. BOOK SECOND. OF THE INTERSECTION OF PLANES. DEFINITIONS. I. A straight line is perpendicular, or at right angles to a plane, when it makes right angles with every straight line which it meets in that plane. II. A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common section of the two planes are perpendicular to the other plane. III. The inclination of a straight line to a plane is the acute angle contained by that straight line, and another drawn from the point in which the first line meets the plane, to the point in which a perpendicular to the plane, drawn from any point of the first line, meets the same plane. IV. The angle made by two planes which cut one another, is the angle contained by two straight lines drawn from any, the same point in the line of their common section, at right angles to that Îine, the one in the one plane, and the other in the other. Of the two adjacent angles made by two lines drawn in this manner, that which is acute is also called the inclination of the planes to one another.* V. Two planes are said to have the same, or a like inclination to one another, which two other planes have, when the angles of inclination above defined are equal to one another. VI. A straight line is said to be parallel to a plane, when it does not meet the plane, though produced ever so far. VII. Planes are said to be parallel to one another, which do not meet, though produced ever so far. VIII. A solid angle is an angle made by the meeting of more than two plane angles, which are not in the same plane, in one point.* * See Notes. PROP. I. THEOR. One part of a straight line cannot be in a plane and another part above it. If it be possible, let AB, part of the straight line ABC, be in the plane, and the part BC above it; and since the straight line AB is in the plane, it can be produced in that plane (I. Post. 2), let it be produced to D. Then ABC and ABD are two straight lines, and they have the common segment AB, which is impossible (I. Def. 3, Cor.) Therefore, ABC is not a straight line. Wherefore, one part, &c. Q. E. D. B Any three straight lines which meet one another, not in the same point, are in one plane. E Let the three straight lines AB, CD, CB meet one another in the points B, C, and E; AB, CD, CB are in one plane. Let any plane pass through the straight line EB, and let the plane be turned about EB, produced, if necessary, until it pass through the point C. Then, because the points E, C are in this plane, the straight line EC is in it (I. Def. 5): for the same reason, the straight line BC is in the same; and, by the hypothesis, EB is in it; therefore the three straight lines EC, CB, BE are in one plane; but the whole of the lines DC, AB, and BC produced, are in the same plane with the parts of them EC, EB, BC (II. Sup. 1). Therefore, AB, CD, CB, are all in one plane. Wherefore, &c. Q. E. D. C B COR. It is manifest, that any two straight lines which cut one another are in one plane. Also, that any three points whatever are in one plane. PROP. III. THEOR. If two planes cut one another, their common section is a straight line. B Let two planes AB, BC cut one another, and let B and D be two points in the line of their common section. From B to D draw the straight line BD; and because the points B and D are in the plane AB, the straight line BD is in that plane (I. Def. 5): for the same reason, it is in the plane CB; the straight line BD is therefore common to the planes AB and BC, or it is the common section of these planes. Therefore, &c. Q. E. D. PROP. IV. THEOR. If a straight line stand at right angles to each of two straight lines in the point of their intersection, it will also be at right angles to the plane in which these lines are.* E B G H Let the straight line AB stand at right angles to each of the straight lines EF, CD in A, the point of their intersection; AB is also at right angles to the plane passing through EF, CD. Through A draw any line AG in the plane in which are EF and CD; take D any point in CA, and make AF-AD; join FD, meeting AG in G; and bisect FD in H (I. 10), and join AH, BD, BG, BH, BF. Then, because AD-AF and AB is common to the two triangles DAB, FAB, and the angle DAB equal to the angle FAB (each of them being a right angle); the side DB is equal to the side FB (I. 4). And because DH-HF and HA common to the triangles DHA, FHA and the base DA=FA; the angle DHA-FHA (I. 8); therefore each of them is a right angle. Similarly each of the angles DHB, FHB is a right angle. Hence, in the triangle DGB, DB2+ 2DG.GH=DG2+GB2 (II. 12). But DB2 DA2+AB2 (I. 47); therefore DA2+AB2+2DG.GH=DG2+GB2. And in the triangle DGA, DA2+2DG.GH=DG2+GA2 (II. 12), therefore DG2+GA2+ AB2+2DG.GH=DG2+GB2+2DG.GH, and by taking away the common part DG2+2DG.GH, we have GA2+AB2=GB2; therefore (I. 48) the angle GAB is a right angle. In the same manner it may be shown that AB is at right angles to any other straight line in the plane of the lines AD, AF; it is therefore at right angles to the plane itself. Therefore, &c. Q. E. D. PROP. V. THEOR. D If three straight lines meet all in one point, and a straight line stand at right angles to each of them in that point; these three straight lines are in one and the same plane. Let the straight line AB stand at right angles to each of the straight lines BC, BD, BE in B, the point where they meet; BC, BD, BE are in one and the same plane. If not, let BD and BE, if possible, be in one plane, and BC be above it; and let a plane pass through AB, BC, the common section of which, with the plane in which BD and BE are, shall be a straight.(II. Sup. 3) line; let this be BF; therefore the three straight lines AB, BC, BF are all in one plane, viz., that which passes B through AB, BC; and because AB stands at right angles to each of the straight lines BD, BE, it is also at right angles (II. Sup. 4) to the Ε plane passing through them; and therefore makes right angles * See Notes. |