D F A B L D EAN B к M 3. Vertical 'angle, 'sum of sides and taining sides, that is, they are equal to area; one another; therefore AEF is the tri4. Base, sum of sides, and area, angle required. 1. Let AB 4. Let A B be the given base, and let be the given it be bisected in C. Let twice CE be base, AC the the given sum of the sides, and let the given sum of triangle D A B contain an area equal to the two sides, the given area. and D the Take CF (II. 52.) a third proportional given vertical to CA, CE, and from the point F draw FG angle. Upon perpendicular to EF (1.44.): through AB (60.) describe a segment A EB, containing an angle equal to the half of D. With the centre X and radius AO describe a circle cutting the arc A EB in E. Join A E, B E, and at the point f B make the angle EBF equal to BEA: the triangle F A B shall be the triangle required. For, because the angle FEB is equal to FBE, the side F B is equal to FE (1.6.), and the two sides AF, FB together are equal to AE, that is, to AC, D draw DG parallel to AB (I. 48.) to the given sum of the sides. Again, be, meet FG in G, and join G A. Through cause the angle A FB (I. 19.) is equal E and C draw E H and C K parallel to to the sum of the angles at E and B, FG, to meet GA and GA produced in and that these angles are equal to one the points H, K, so that KG, KH, KA another, the angle A F B is equal to will be proportionals (II. 29.): from the twice the angle at E, that is, to the given centre K, with the radius KH, describe a angle D. And the triangle is described circle cutting G D in L, and join L A, upon the given base AB. LB: LAB shall be the triangle required. Therefore, &c. Produce G K to M, so that KM may 2. Let A B be be equal to KH; and from L draw LŇ the given base, perpendicular to A B. Then, because upon which let. KÁ, KH, K G are proportionals, MA, there be describ MH, MG are in harmonical progression ed the rectangle À BCD, contain (II. 46.); but the point L is in the cir cumference of a circle upon the mean ing an area equal to twice the given MH; therefore (51. Cor.) LA : LG:: area (1. 57.), and a segment AEB con AH: HG; but LG is (I. 22.) equal taining an angle equal to the given an to NF, and AH:HG::AE: EF gle (60.). Then if the arc AEB cut the side CD in E, and EA, EB be (II. 29.), that is, since CA, CE, CF are joined, E A B will evidently be the tri- CE ; therefore L A is to N F as CA to proportionals (II. 22. Cor. 1.)::CA: angle required. CE, or in the subduplicate ratio of CA 3. Let A be the to CF; and (II. 38. Schol. Lem. 1. Cor.) given vertical an if B f be taken equal to A F, L B is to gle, and let the tri Nf in the same ratio. Therefore the angle ABC (11.69.) sum of L A, L B is to the whole line contain Ff in the same ratio, or, if Ce be taken equal to the given equal to CE, as E e to Ff: therefore, area: and let D be the sum of LA, LB is equal (II. 11. the given sum of the Cor. 1.) to E e, that is, to the given sum, two sides. Divide and LAB is the triangle required. D into two parts, Therefore, &c. * such that their rect.angle may be equal to the rectangle un If the difference of the sides be supposed given der AB, AC (11.56.). Take AE equal to instead of the sum in cases 1, 3 and 4, solutions of the same character may be obtained ; viz. in case 1, one, and AF equal to the other of these by describing upon the given base a B a segment parts, and join E F. Then, because the which shall contain an angle exceeding by a right triangles ABC, AEF have the common angle half the given angle B; in case 3, by dividing angle A, they are to one another (II.40. under the segments may be equal to A B XA C; and the difference D produced, so that the rectangle Cor.) as the rectangles under the con- in case 4, by making use of the following corollary to D B an area B B and C D in F, and take EG a third proPROP. 65. Prob. 12. portional to EB and C F. From the To find two straight lines, there being centre E, with the radius E A, describe assumed any two out of the six follow- a semicircle AHB; and from the point ing data ; viz. their sum, their differ- G (I. 44.) draw G H perpendicular to ence, the sum of their squares, the dif- A B, to meet the circumference in H. ference of their squares, their ratio, and Join AH, HB: they shall be the their rectangle. straight lines required, given sum of the Bisect and from D (1.44.) draw DE per- pendicular to AC. From the centre D 6. Difference,and difference of squares. circle cutting D E in E; and from the with the radius D A or DC describe a centre E, with the radius EA or EC 7. Ratio, and rectangle (II. 63.) describe the circle CFA: lastly, from 8. Sum, and ratio (II. 55. fig. 1.) the centre A with the radius A B de9. Difference, and ratio (II.55. fig. 2.) scribe an arc cutting CFA in F. Join 10. Sum, and rectangle (II. 56. fig. 1.) AF, let AF cut the circumference 11. Difference, and rectangle (11. 56. AEC in G: and join G C. AG and fig. 2.) GC shall be the lines required. 4. Let A B be the given difference, 12. Sum of squares, and ratio. and the square of A C the given sum of 13. Difference of squares, and ratio. the squares. Bisect 14. Sum of squares, and rectangle. AC in D (I, 43.), and 15. Difference of squares, and rectan- from D (1.44.) draw gle. DE perpendicular to AC: from the centre DA or DC describe a circle cutting DE centre E with the radius EA or EC de They will accordingly be here omitted. Of the rest the greater part are so ob. scribe the circle CFA; lastly, from the vious, that it will be sufficient to indi- centre A with the radius AB describe cate only the construction, leaving the let AF produced cut the circumference an arc cutting CFA in F: join AF, and demonstration to the reader. 1. Let AB be the given sum, AC the AECG in G, and join GC. AG and given difference. Bisect C B in D, and GC shall be the straight lines required. 5. Let A B be the given sum, and the square of A C the given difference of the squares: take A D a third pro- 오면 portional to AB, AC (II. 52.), and CD2 the given bisect D B in E (I. 43.); AE, EB difference of the shall be the straight lines required. squares. Bisect 6. Let A B be the given difference, AB in E (1. 43.) D A B А C D B B H the squares, and E .A. G B a bisect B D in E (1.43.) ; A E, E B shall upon A D_describe a circle cutting BE be the straight lines required. in E: A E, EB shall be the straight 12. Let the given ratio be that of lines required. For it is evident that A B to A C, and let the square of D be the square of ED being (as in the last the given sum of the case) by the construction equal to the squares: from A (I. rectangle under AD, DB (I1. 34. Cor.), 44.) draw A E at right ED is equal to C; and, because the triangles to A B, and from angles ABE, EBD are similar (II. 34.), the centre A with the AB is to AE as EB to ED, and there. radius AC describe a fore the rectangle under AE, EB is circle cutting A E in E: (II. 38.) equal to the rectangle under join B E: take B F equal AB, ED, that is to the rectangle under to D, and through F AB and C. (I. 48.) draw FG parallel to EA: GF, Scholium. G B shall be the straight lines required. 13. Let the given ratio be that of In the cases of this Proposition we AB to AC, and let the square of D have exhibited geometrical solutions of be the given difference of the squares: the following well-known equations. from B (1.44.) draw 1. x +y =a 2. 2? +yo = a BE at right angles x - y = 6 x? - y = 62 to AB, and from the centre A, with the 3. x +y =a 4. x - y =a radius A C, describe x+y = x2 + y2 = 62 a circle cutting BE 5. x +y =a 6. x - y =a in E: join A E: take B F equal to D, and X®- y2 = 62 2® +y = through F (I. 48.) draw FG parallel to AE: GF, GB shall be the straight 7. x lines required. 14. Let the square of A B be the given sum of the squares, and let the 8. x+y= a rectangle under A B and C be the 9. x - y = a 6 given rectangle : divide A B in D (II. y 56.), so that the rect 10. x +y = a 11. X angle under AD, DB may be equal to the square of C; from D(I1.56.) draw DE 12. Qe+ y2 = a 13. x2 - y2 =a? at_right angles to 6 6 AB; and upon AB as a diameter de y y scribe a circle cutting DE in E: AE, EB shall be the straight lines required. 14. 22 + y2 =a 15. x2 - y2 = a? For it is evident, that the square of ED XY = 68 being, by construction, equal to the rect- In the construction of these and other angle under AD, DB (II. 34.Cor.), ED is problems of the foregoing Sections, the equal to C; and the rectangle under data have always been supposed such AE, EB is equal to the rectangle under that the problem in question be not imAB, ED), that is, to the rectangle under possible. For, as we have already had AB and C. occasion to observe, many of them are 15. Let the square of A B be the possible, only so long as the mutual given difference of the squares, and let relations of the data are confined within the rectangle under A B and C be the certain limits. Thus, if it be required given rectangle; to find two lines, such that their squares produce AB to may together contain 9 square feet, it D (II. 56.), so is evident that the sum of the lines in that the rectangle question must not be less than 3 feet, under AD, DB nor must their difference exceed 3 feet, may be equal to (11.56. N.B.). The solution, therefore, the square of C, of a problem, which should require the from B (I. 44.) sum of the two to fall short of this quandraw B E at right angles to AD, and tity, or their difference to exceed it, ху abc с xy = 62 A D с B A would be impossible. In this manner 2. (Euc. xi. def. 5.) A straight line is does one of the conditions frequently set said to be inclined to a plane, when it limits to the other-frequently, but not meets the plane but is not perpendicular in every case :-thus, if two lines be re- to it. quired, which shall contain a given When a straight line A B is inclined rectangle, their ratio may be any what- to a plane CDE, ever, and a problem which should re- the angle ABF A quire them to be to one another in any which it makes ratio, how great or how small soever, with a straight would be possible. The limits of possi- line drawn from bility, when there are any, are commonly the point B in indicated by the construction, if the which it meets problem be solved geometrically, as the plane, through D D they are, if algebraically, by the form the foot of the of the final equation. See the cases of perpendicular AF, which is let fall upon Prop. 64., where the vertex of the tri- the plane from any other point A of the angle sought is determined by the inter- straight line (see Prop. 7.), is called the section of a straight line and circle, or of angle of inclination. two circles : if the data be such that no 3. A straight line is said to be paintersection can take place, the con- rallel to a plane, when it cannot meet struction fails, and the problem becomes the plane, to whatever extent both be impossible. produced. Also, conversely, in this case BOOK IV. the plane is said to be parallel to the $1. Of Lines perpendicular, or inclined, straight line. or parallel to planes.-§ 2. Of Planes 4. If two planes ABC, which are parallel, or inclined, or per- ABD intersect one anpendicular to other Planes.—5 3. Of other in a line as A B Solids contained by Planes. —$ 4. (see Prop. 2.), they are Problems. said to form at that line SECTION 1.—Of Lines perpendicular, a dihedral angle CABD. or inclined, or parallel to Planes. The magnitude of a dihedral angle does not In the preceding books our attention has been confined to lines which lie in of the containing planes , but upon the depend upon the extent one and the same plane, the intersection of such lines, and ihe figures contained hedral angle C ABD is greater than opening between them. Thus, the diby them; we are now to consider lines which lie in different planes, planes dral angle C A BE. the dihedral angle E ABD by the dihewhich intersect one another, and solids which are contained by plane or other another plane makes the adjacent dihe 5. When one plane standing upon surfaces. In other words, we have dral angles equal to one another, each been hitherto engaged with Plane Geo of them is called a metry; we are now to enter upon Solid right dihedral angle; Geometry. Det. 1: (Euc. xi. def.3.) A straight line and the plane which is said to be perpendicular (or at right is said to be perpen stands the other angles) to a plane, when it makes right dicular (or at right angles with every angles) to it. straight line meeting it in that plane, A dihedral angle is also said to be (see Prop 3.). Also, acute or obtuse, according as it less or conversely, in this greater than a right angle. case the plane is 6. (Euc. xi. def. 8.) Planes, which do said to be perpendi not meet one another, though produced cular to the straight to any extent, are said to be parallel. line. 7. (Euc. xi. def. 9.) If three or more The foot of the perpendicular is the planes pass through a point as A, they are point* in which it meets the plane. * It is evident that a straight line cannot meet a face common to both, unless they coincide altogether. plane in more than one point, unless it lies alto- (See Prop. 1.) Therefore a straight line cuts a plane gether in the plane; and in like manner that one in a point ; and a plane cuts a plane in a line, which plane cannot meet another plane in a portion of sur- line (see Prop. 2.) is a straight line. D E B a D B $ said to form at that 11. (Euc. xi. def. A.) A parallelopiped point a solid angle, of is a solid figure hav A which the intercepted ing six faces, of which plane angies (see every opposite two Prop. 2.) are called are parallel. the sides or faces, and A Such a figure mayA the intersections of be formed by taking the planes, edges. any solid angle A, which is formed by 8. A polyhedron is a solid figure in- B, C, D in the three edges, and passing three plane angles, assuming any points cluded by any number of planes, which through those points planes parallel to are called its faces : if it have four faces the planes A Č D, A B D, A B C reonly, which is the least number possible, it is called a tetrahedron ; if six, a spectively. hexahedron ; if eight, an octahedron ; sometimes distinguished by naming any The faces of a parallelopiped are if twelve, a dodecahedron; if twenty, an icosahedron; and so on.* two opposite faces the bases of the paThe intersections of the faces of a rallelopèped, and the other four the sides : in which case the altitude of the polyhedron are called, arrises or edges, parallelopiped is the perpendicular disand the points of the solid angles tance between he two bases. vertices or angular points. The diagonals of a polyhedron are the straight lines that which has one of 12. A rectangular parallelopiped is which join any two vertices not lying in its solid angles con. the same face.. The surfaces of the polyhedrons here tained by three right treated of are supposed to be convex, (see Prop. 17.) every angles, and therefore that is, such that the same straight line face at right angles to can cut them in two points only. those which are adjoining to it. 9. A polyhedron is said to be regular, when its faces are similar and equal 13. A cube is a rectangular paralleloregular polygons, and its solid angles piped which has the three equal to one another. There are only the solid angles equal to edges terminated in one of five such figures. (See Prop. 20. Cor.) one another. The cube of any straight line A B, is 14. (Euc. xi. def. 13.) A prism is a solid figure having any number of faces, two of which are similar and equal rectilineal figures, so placed as to have their corresponding sides parallel, and the rest parallelograms. lar faces similarly situated, and forming Such a figure may be formed by equal dihedral and solid angles." * The Greek word for “ seat” being in all cases 2. If all the planes but one which form a convex annexed to the Greek numeral which indicates the surface be similar and similarly situated to all the number of seats, or faces, on which the figure may be planes but one which contain a solid figure, each to A solid figure may be contained by any each, and if the dihedral angles which every adjoinnumber of faces above three, in the same manner as ing two of the first make with one another be equal to the dihedral angles which every corresponding a plane figure may be contained by any number of sides above two, the numbers 4, 6, 8, 12, and 20 being two of the latter make with one another, each to here specitied unly because the other solids (of 5, 7, each, the remaining edges of the surface (viz. those &c. faces) are less frequently subjects of considera- which are not common to adjoining planes) shall tion. lie in one plane, and shall inclose a rectilineal figure + The same observation may be made here as at similar to the last face of the solid, and making Book 11. def. 14, viz. that in this definition there are equal dihedral angles with the corresponding faces some things assumed which have not been as yet adjoining to it. This may be demonstrated by demonstrated. These are, making any two of the equal solid angles coincide. 1. If all the plane angles but one which contain (See Prop. 14. Cor.) It is evident, also, that the definition would be two solid angles be equal, each to each, in order, and make with one another equal dihedral angles, the complete without mentioning the equality of the remaining plane angle of the one shall be equal to solid angles, for the several plane and dihedral anthe remaining plane angle of the other, and the two gles of the one being equal and similarly situated remaining dibedral angles of the one equal respec- with the corresponding plane and dihedral angles of the other, it is evident that any two corresponding tively to the two remaining dihedral angles of the other. This may be proved by coincidence. solid angles may be said to coincide. seated. |