Secondly. The altitude CM being common, the triangle CPM is E P. M Q to the triangle CPE as PM is to B PE; but since CP bisects the an D gle MCE, we have PM : PE :: CM : CE (Book IV. Prop. XVII.) :: CD ; CA::A:A: hence CPM : CPE ::A:A'; and consequently CPM : CPM+ CPE or CME :: A:A+A'. But CMPA, or 2CMP, and CME are to each other as the polygons B' C and B, of which they form part : hence B': B :: 2A : A+A'. Now A' has been already determined ; this new proportion will 2A.B serve for determining B', and give us B'= and thus by A+A means of the polygons A and B it is easy to find the polygons. A' and B', which shall have double the number of sides. PROPOSITION XIV. PROBLEM. To find the approximate ratio of the circumference to the diameter. 16 Let the radius of the circle be 1; the side of the inscribed square will be V2 (Prop. III. Sch.), that of the circumscribed square will be equal to the diameter 2 ; hence the surface of the inscribed square is 2, and that of the circumscribed square is 4. Let us therefore put A=2, and B=4 ; by the last proposition we shall find the inscribed octagon A'=V8=2,8284271, and the circumscribed octagon B=2+18=3.3137085. The inscribed and the circumscribed octagons being thus determined, we shall easily, by means of them, determine the polygons having twice the number of sides. We have only in this case to put A=2.8284271, B=3.3137085; we shall find A'= 2 A.B A.B=3.0614674, and B'= A+A=3.1825979. These polygons of 16 sides will in their turn enable us to find the polygons of 32; and the process may be continued, till there remains no longer any difference between the inscribed and the circumscribed polygon, at least so far as that place of decimals where the computation stops, and so far as the seventh place, in this example. Being arrived at this point, we shall infer that the last result expresses the area of the circle, which, since it must always lie between the inscribed and the circumscribed polygon, and since those polygons agree as far as a certain place of decimals, must also agree with both as far as the same place. We have subjoined the computation of those polygons, carried on till they agree as far as the seventh place of decimals. The area of the circle, we infer therefore, is equal to 3.1415926. Some doubt may exist perhaps about the last decimal figure, owing to errors proceeding from the parts omitted; but the calculation has been carried on with an additional figure, that the final result here given might be absolutely correct even to the last decimal place. Since the area of the circle is equal to half the circumference multiplied by the radius, the half circumference must be 3.1415926, when the radius is 1; or the whole circumference must be 3.1415926, when the diameter is 1 : hence the ratio of the circumference to the diameter, formerly expressed by , is equal to 3.1415926. The number 3.1416 is the one generally used. BOOK VI. PLANES AND SOLID ANGLES.. Definitions 1. A straight line is perpendicular to a plane, when it is perpendicular to all the straight lines which pass through its foot in the plane. Conversely, the plane is perpendicular to the line. The foot of the perpendicular is the point in which the perpendicular line meets the plane. 2. A line is parallel to a plane, when it cannot meet that plane, to whatever distance both be produced. Conversely, the plane is parallel to the line. 3. Two planes are parallel to each other, when they cannot meet, to whatever distance both be produced. 4. The angle or mutual inclination of two planes is the quantity, greater or less, by which they separate from each other; this angle is measured by the angle contained between two lines, one in each plane, and both perpendicular to the common intersection at the same point. This angle may be acute, obtuse, or a right angle. If it is a right angle, the two planes are perpendicular to each other. 5. A solid angle is the angular space included between several planes which meet at the same point. Thus, the solid angle S, is formed by D the union of the planes ASB, BSC, CSD, с DSA. Three planes at least, are requisite to form a solid angle. A B PROPOSITION I. THEOREM. A straight line cannot be partly in a plane, and partly out of it. For, by the definition of a plane, when a straight line has two points common with a plane, it lies wholly in that plane, Scholium.' To discover whether a surface is plane, it is necessary to apply a straight line in different ways to that surface, and ascertain if it touches the surface throughout its whole extent. PROPOSITION II. THEOREM. Two straight lines, which intersect each other, lie in the same plane, and determine its position, Let AB, AC, be two straight lines which intersect each other in A; a plane may. be conceived in which the straight line AB is found ; if this plane be turned round AB, until it pass through the point C, then the line AC, B which has two of its points A and C, in this plane, lies wholly in it; hence the position of the plane is determined by the single condition of containing the two straight lines AB, AC. Cor. 1. A triangle ABC, or three points A, B, C, not in a straight line, determine the position of a plane. If two planes cut each other, their common intersection will be a straight line. E Let the two planes AB, CD, cut C each other. Draw the straight line EF, joining any two points E and Fin A the common section of the two planes. This line will lie wholly in the plane AB, and also wholly in the plane CD (Book I. Def. 6.): therefore it will be in both planes at once, and consequently is their common intersection. BI F D PROPOSITION IV. THEOREM. If a straight line be perpendicular to two straight lines at their point of intersection, it will be perpendicular to the plane of those lines. Let MN be the plane of the A two lines BB, CC, and let AP be perpendicular to them at their point of intersection P; then will AP be perpendicular M to every line of the plane pass B В ing through P, and cor.sequently to the plane itself (Def. 1.). P Q Through P, draw in the plane с MN, any straight line as PQ, B and through any point of this N line, as Q, draw BQC, so that BQ shall be equal to QC (Book IV. Prob. V.); draw AB, AQ, AC, The base BC being divided into two equal parts at the point Q, the triangle BPC will give (Book IV. Prop. XIV.), PC?+PB2=2PQ2+2QC?. The triangle BAC will in like manner give, AC + AB2=2AQ”+2QC. Taking the first equation from the second, and observing that the triangles APC, APB, which are both right angled at P, give AC?_PC2=AP?, and AB -PB’=AP; we shall have AP2 + AP2=24Q2_2PQ?. Therefore, by taking the halves of both, we have AP=AQAPQ?, or AQ?=AP2+PQ2 ; hence the traiangle APQ is right angled at P; hence AP is per pendicular to PR. |