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Let A: B::

A:mB::nA : nB, &c.

Then will A: B: : A + mA + nA: B+ let BK be perpendicular to AE, diaw CK, FL pa-
BnB, &c.
B+mBnB B
= the fame ratio.
A+MA+NA A

For

THEOR. XXXII. If a whole magnitude be to a whole, as a quantity taken from the first to a quantity taken from the other, the remainder is to the remainder, as the whole is to the whole.

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Then is A X mB = B × mA ≈ mAB, as is evident.

COR. If three quantities be continual proportionals, the rectangle or product of the extremes will be equal to the fquare of the mean.

THEOR. XXXV. fig. 51. Triangles and alfo parallelograms having the fame altitude, or that are between the same parallels, are to one another in the fame ratio as their bafes.

Let the triangles ADC, DEF have the fame altitude, or be between the same parallels AE, CF, then AD: DE :: Triangle ADC: Triangle DEF. For let AD be to DE as any one number m (2) to any other number n (3), and divide the bafes into parts AB, BD, DG, GH, HE all equal, and join BC, FG, FH; thefe lines will evidently divide the triangles into the fame number of parts as their bafes, each equal to the triangle ABC (Th. 16. Cor. 1.) fo that the triangle ADC contains m (2) fuch equal parts as the triangle DEF contains n(3); now the base AD also contains m (2) such equal parts as the bafe DE contains 7 (3); therefore the triangle ADC is to the triangle DEF as m to n, or as AD to DE. In like manner the parallelogram ADKI is to the parallelogram DEFK as the base AD to the base DE.

equal bases AB, BE in the fame ftraight line AE; rallel to AE, and join AK, AL. The triangles ACB, AKB are equal, alfo the triangles ALB, BFE. (h. 16. Cor. 3.) And by the theorem the triangle AKB is to the tringle ALB as KB to LB, but KB and LB are equal refpectively to CG and FH the altitudes of the triangles ACB, BFE; therefore the triangles ACB, BFE are to each other as their altitudes.

THEOR. XXXVI. fg. 53. If a line be drawn in a triangle paralel to one of its fides, it cuts the two other fides proportionally.

Let DE be drawn parallel to BC ore of the fides of the triangle ABC, then AD: DB:: AE: EC. Join BE and CD. The triangles BDE, CDE are equal (Th. 16, Cor. 1.) therefore, fince any magnitude muft neceflarily have the tame ratio to cach of two equal magnitudes, the triangle ADE is to the triangle BDE as the fame triangle ADE to the triangle CDE, but ADE is to EDB as AD to DB (Th. 35.) and ADE is to EDC as AE to EC, therefore AD: DB::AE: CE.

COR. 1. Hence alfo the whole lines AB, AC are proportional to their proportional fegments (Cor. Th. 25), namely AB: AC:;AD: ÀE and

AB: AC:: BD: CE.

COR. 2. A ftraight line which divides the fides of a triangle proportionally is parallel to the remaining fide.

THEOR. XXXVII. fg. 54. If a line bifect the vertical angle of a triangle and meet the bafe, the fegments of the bafe will be directly proportional to the other two fides of the triangle.

In the triangle ABC let BD bife&t the vertical angle, then AB: BC:: AD: DC. For, draw CE parallel to DB, meeting AB produced in E. The angle ECB is equal to CBD (Theor. 10.) er to DBA (by hyp.) that is to CFA or CEB (Th. 10. Cor.) therefore CB is equal_to BE (Th. 4.) In the triangle AEC, the line BD is parallel to EC, theretore AB: BE; AD: DC, or fince BE is equal to BC, AB: BC::AD: DC.

COR. Hence, converfely, if a line be drawn from the vertex of a triangle to divide the bafe into fegments directly proportional to the fides, that line will bifect the vertical angle.

THEOR. XXXVIII. fig. 55. Equiangular tri'angles are fimilar, or have their like fides propor. tional.

COR. fig. 52. Triangles and parallelograms having equal bafes are to one another as their alti-; tudes. For let ABC, BFE be triangles having their

Let ABC, DEF be two equiangular triangles having the angle A equal to the angle D, the angle B to the angle E, and confequently the angle C to the angle F, then will AB: AC:: DE: DF. Take DG equal to AB and DH to AC and join HG. Then the triangles ABC, DGH are equal in all refpects, (Th. 1.) namely the angles Band C equal to the angles G and H, but B and C are equal to E and F (by hyp.), therefore G and H are equal to E and F, and confequently G H is parallel to E F, (Th. 11. Cor.), therefore DG DH:: DE: DF (Th. 36. Cor. 1.) but DG and DH are equal to AB and AC, therefore AB : AC :: DE: DF.

THEOR. XXXIX. fig. 55. Triangles which have their fides proportional are equiangular. In the triangles ABC, DFF, if AB: AC :: DE DF:: BC: EF, the triangles will have their Y y a

correfponding angles equal. Take DH equal to AC and DG equal to AB. Then DG: DH:: DE: DF, therefore GH is parallel to EF (Th. 36, Cor. 2); hence the triangles DGH, DEF are equiangular (Th. 10.); wherefore DG: GH:: DE: EF (Th. 38.):: AB : BC (by hyp.); fince there. fore DG GH:; AB: BC, and that DG is equal to AB, therefore GH is equal to BC. Thus the triangles DG, ABC, having the three fides of the one refpectively equal to the three fides of the other, are cquiangular (Th. 5.) therefore alfo the triangles ABC, DEF are equiangular.

THEOR. XL. f. 55. Triangles which have one angle in the one equal to one angle in the other, and the fides about thefe angles proportional, are equiangular.

Let ABC, DFF be two triangles having the angles A and D equal, and AB: AC::DE: DF; thefe triangles fhall be equiangular. Make DG equal to AB, and DH to AC, and join GH: thus the triangles ABC, DG!! are identical and equiangular (Th. 1.); therefore HD: DG :: CA: AB:: FD: DE (by hyp.); therefore HG is parallel to FE, (Th. 36. Cor. 2.) and the triangles HDG, FDE, afo CAB, FDE are equiangular.

the angles AED, CEB are oppofite (fig. 57), and therefore equal (Th. 7.): or the angle at E is coramon to both triangles (fig. 58.), in either cafe the triangles are equiangular; therefore DE: EA:: EB: EC (Th. 38.); hence the rectangle of DE, EC is equal to the rectangle of AE, EB, (Th. 41.) COR. If the line BAE. (fig. 58.) be fuppofod by revolving to come into the pofition of the tangent AE (g. 59), the diftances BE, AE will thus have become equal. Hence we have this THEOREM. If from a point without a circle two lines be drawn, one touching it, and the other cutting it, the rectangle of the diftances of that point from the interfections of the cutting line, or fccant, is equal to the fquare of the tangent.

THEOR. XLIII, fig. 69. In a right ang ed triangle, a perpendicular from the right angle is a mean proportional between the fegments of the hypothenufe; and each of the fides about the right angle is a mean proportional between the adjacent fegment, and the hypothenufe.

Let ABC be a rigns angled triangle, and CD a perpendicular upon the hypothenu; then will AD; DC :: DC: DB, and AB : AC :: AC:AD,

and AS: BC :: PC: BD.

For the tringles ACB, ADC having the right

"THEOR. XLI.fig, 56. If four lines are proportional, the rectangle of the extremes will be equal`angtis ** C a d D equal, and the angle at A com. to the rectangle of the means; and if the rectangle of the extremes be equal to the rectangle of the means, the four lines are proportional.

Let the four lines A, B, C, D be proportion, or A: B::C:D, then will the rectangle of A and D be equal to the rectangle of B and C. Let the four lines be placed with their extremitics meeting at a common point, and forming four right angles; and draw lines parallel to them to complate the rectangles P, Q, R; where P is the rectangle of A and D, Q the rectangle of B and D, and R the rectangle of B and C. Then the acctangles P and Q will be to each other as A and B (Th. 35.) and in like manner the rectangles R and Q will be to each other as C and D; but the ra. tio of A to B is the fame as the ratio of C to D; therefore the ratio of P to Q is the fame as the ratio of R to Q, and confequently P and Rare equal. Again, if the rectangle of A and D be equal to the rectangle of B and C, A: B::C: D. For the rectangles being placed as before, it is evident that P and R have each the same ratio to Q; but Pis to Qas A to B, and R to Q as Cto D, theretore A: B:: C: D,

on, have their tind angles equal, and are equiangulor; and in like manner it will appear that the triangle. ACB, CDB are equiangular. Hence these the triangles ACB, ADC, CDB being equiangular, will have the fides about the equal angles propo.tional; thus we get AD : DŮ :: DC: DB, and AB: AC :: AC : AD, and AB ; BC:: "C:PD (Th_38)

THEOR, XLIV. fg 61. Fquiangular or fimilar triangles are to each, ther as the fquares of their like fides.

Let ABC, DEF, be two equiangular triangles, AB and DE being their homologous or like fide, and AL DW iquares on these fides. The triang ABC is to the triangle DEF as the fquare AL to the fquare DN. Draw CG and FH perpendicu lar to AB and DF, and join BK and EM. The triangles ACG, DFH are equiangular (Theor. 12, Cor. 3.); therefore AC : DF ::CG:FH(Th. 38.); but the triangles ABC, DEF being equiangular, we have AC: DF:: AB: DE; therefore, from equality of ratios, we have CG : FH :: AB : DE:

AK: DM, and by alternation, CG: AK :: FH : DM. Now CG : AB ; : triangle ABC : tri. ABK (Th. 35, Cor.); and in like manner FH: DM :

Coa. If three lines are proportional, the rect. angle of the extremes is equal to the fquare of the triangle DFE: DME, therefore tri. ABC : tri. nean; and if the rectangle of the extremes be e- ABK :: trl. DFE : tri. DME, and by alternation, qual to the fquare of the mean, the three lines are tri. ABC: tri. DFE : : tri. ABK : trì. DME. Pat proportional. the fquares AL. DN being the doubles of the triTHEOR. XLII. fg. 57 and 58. If two lines meet-angles ABK, DFE, have the fame ratio with ing a circle cut each other, either within it, or without, the rectangle of the parts of the one will be equal to the rectangle of the parts of the other; the parts of each being measured from the point of meeting to the two interfections with the circumference.

Let the two chords, AB, CD, m et each other in E, the rectangle of AE, EB is equal to the rectangle of CE, ED. Join AD, and CB. The Timangles AED, CEB are equiangular, for the anges at D and B are equal (Th. 21. Cor. 1., and

them: Therefore the triangle ABC is to the triangle DFE, as the fquare AL to the fquare DN.

THEOR. XLV. fig. 62. Similar rectilineal figures are to each other as the fquares of their like fides.

Let ABCDE, FGHIK be two fimilar figures, the like fides being AB and FG, BC and CII, and fɔ on; the figure ABCDE will be to the fure FGIK as the fquare of AB to the fquare of Join BE, BD, GK, CI. Because the angle A and F are equal and BA : AE :: GF; FK, the triangles BAE, GFK are equiangular (11,4^

hence AE: EB:: FK: KG, but AE: ED: : FK :KI (by hyp.), therefore EE: ED :: CK: KI. Now the angles AED, FKI are equal, and the angles AEB, FKG have been proved equal; therefore the angles BED, GKI are equal; thus the triangles BED, GKI are alfo equiangular, and in the fame way it may be fhewn that the triangles BDC, GIH are equiangular. The triangle ABE is to FGK as the fquare of BE to the fquare of GK, that is, as the triangle FBD to the triangle KGI (Th. 44.), and in like manner it will appear, that EBD is to KGF as DBC to IGH: Therefore the whole figure ABCDE is to the figure FGHIK, as the triangle ABE to the triangle FGK (Th. 31.); that is, as the fquare of AB to the iquare of FG (Th. 44.).

SCHOLIUM. From this propofition it may be demonftrated, that circles are to one another as the fquares of their diameters And in general, that all fimilar plane figures whatever, are to one another as the fquares of their like parts.

For let ABCDEF GHKLMN, (fig. 63.) be any two regular polygons, of the fame number of fides, infcribed in circles whofe diameters are AD, GL. Draw AO, FO to the centre of the one polygon, and GP, NP to the centre of the other. The angles AOF, GPN, standing each upon the fame part of the whole circumference, are evidently equal, and confequently the ifofceles triangles, AOF, CPN, are fimilar: Thus it appears that each of the polygons is made up of the fame number of fimilar triangles; therefore the polygon, ABCDEF, is to the polygon, GHKLMN, as the triangle AOF to the triangle GPN; that is, as the fquare of AO to the fquare of GP, or as the fquare of the diameter AD to the fquare of the diameter GL. Now whatever be the number of the fides of the polygon, it is evident, that their proportion to each other will be the fame; namely, that of the fquares of the diameters of their circumferibing circles. By fuppofing the number of the fides of the polygons continually increased, it is evident that their areas will approach more and more to the areas of their circumfcribing circles, which may be confidered as their limits; for it may be demonstrated, that a polygon may have its fides fo numerous as to differ from the area of its circumfcribing circle by lefs than any affignable quantity. Hence we may conclude, that the area of the circles themselves have to each other the fame proportion as their infcribed polygons; namely that of the fquares of the diameters.

SECT. III. Of PLANES and SOLIDS.

DEFINITIONS.

67. THE COMMON SECTION of two planes, is the line in which they meet, or cut each other. 68. A ftraight line is PERPENDICULAR to a plane, when it is perpendicular to every line which ineets it in that plane.

69. One plane is PERPENDICULAR to another, when every right line in the one, which is perpendicular to their line of common section, is perpendicular to the other.

70. The INCLINATION of one plane to another, or the angle they form between them, is the angle contained by two right lines, drawn from any

point in the common fection, and at right angles to the fame, one of thefe lines in each plane.

71. PARALLEL PLANES are fuch as being produced ever fo far both ways, will never meet, or which are every where at an equal perpendicular diftance.

72. A SOLID is that which has length, breadth, and hickness.

73. A PRISM is a folid whofe ends are parallel, equal, and like plane figures; and its fides connecting those ends, are parailelograms. Fig. 64.

74. A PARALLELOPIPED, OF PARALLELOPIPEDON, is a folid bounded by fix parallelograms, every oppofite two of which are equal, alike, and pa, rallel. If the bounding planes are rectangles, it is a RECTANGULAR PARALLELOPIPEDON. Fig. 65. 75. A CUBE is a rectangular parallelopipedon, whofe fix bounding fides are fquares. Fig. 66. 76. A CYLINDER is a folid, conceived to be generated by the revolution of a rectangle about one of its fides, fuppofed to be at reft. The fixed line, about which it revolves, is called its Axis. Fig. 67.

77. A PYRAMID is a folid, whofe bafe is any right-lined figure, and its fides triangles, having all their vertices meeting at a point above the bafe, called the VERTEX of the pyramid. Fig. 68.

78. A CONE is a folid, conceived to be generated by the revolution of a right-angled triangle a bout its perpendicular, which fixed line is called the Axis of the cone. Fig. 69.

79. A SPHERE is a folid defcribed by the revo lution of a femicircle about its diameter; the Exed line, about which it revolves, is called the Axis of the sphere. Fig. 120.

THEOR. XLVI, fg. 70. A PERPENDICULAR IS the fhorteft line that can be drawn from any point to a plane.

1 et AB be perpendicular to the plane DE, then any other line, as AC, drawn from the fame point A to the plane, will be longer than AB. Join BC; then ABC is a right angle, hence BAC is lefs than a right angle, and confequently BA lefs than BC, (Tà. 13.)

COR. A perpendicular measures the diftance of any point from a plane.

THEOR. XLVII, fig. 71. The common fection of two pianes is a traight line.

Let ACBDA, AEBFA, be two planes cutting each other, and A, B two points in which the two planes meet; the straight line joining these points will be the common interfection of the planes, For, because the ftraight line AB touches both planes at the points A, B, it touches them in all other points (Def. 5.); this line is therefore common to both plancs, that is, their common interfection is a ftraight line.

THEOR. XLVIII, fig. 72. If a ftraight line be perpendicular to two other ftraight lines, at their common interfection, it will be perpendicular to the plane of those straight lines.

Let the line AB make right angles with the lines AC, AD, it will be perpendicular to the plane CDE, which paffes through thefe lines. For, if the line AB were not perpendicular to the plane CDE, another plane might pafs through the point A, to which AB would be perpendicular; but this is impoffible, for fince the angles BAC,

BAD,

BAD, are right angles, this other plane must pass through the points C, D. Hence this plane paffing through the points A, C of the line. AC, and allo through the points A, D of the line AD, it will pafs through both thefe lines, and therefore be the fame plane with the former.

COR. If a straight line ftand at right angles to each of three straight lines at the fame point, these three lines are in one plane.

THEOR. XLIX. fig. 73. If two ftraight lines be perpendicular to the fame plane, they will be parallel to each other.

Let AB and CD be both perpendicular to the plane EF; make DG equal to BA, and join Band D, and draw DG perpendicular to BD, in the plane EF; make DG equal to BA, and join AD, AG. The triangles BDG, DBA, have the fides DG, BA, equal, and BD common to both; the angles BDG, DBA are alfo equal, being right angles; therefore thyte triangles are identical, (Th. 1.) hence BG is equal to AD, and the triangles ABG, GDA have two fides AB, BG of the one, equal to two fides GD, DA of the other, each to each, and the fide AG common to both; therefore these allo are identical (Th, 5), hence the angle ADG is equal to BDG, that is to a right angle. Hence it appears that DG is perpendicular to the lines BD, AD; and it is alfo perpendicular to DC; (Def. 68.) Therefore the lines BD, DA, DC are in the fame plane. (Th. 48, Cor.) Since it thus appears that AB, CD, lines in the fame plane, are both perpendicular to a third line BD, the lines AB, CD are parallel. (Th. 8.) COR. If two lincs be parallel, and one of them perpendicular to any plane, the other will alfo be perpendicular to the fame plane.

THEOR. L. fig. 74. If two planes cut each other at right angles, and a ftraight line be drawn in one of the planes, perpendicular to their comnon interfection, it will be perpendicular to the other plane.

Let the planes ACBD, AEBF, cut each other at right angles, and the line CG be perpendicular to their common fection AB; then will CG be perpendicular to the plane AEBF. For, let FG be perpendicular to AB, thus the angle CGF is the angle of inclination of the planes (Def. 70.), and is therefore a right angle; fince therefore the line CG is perpendicular to the two lines AG, GF, it is perpendicular to the plane AEBF, in which thefe lines are drawn. (Th. 48.)

THEOR. LI. fig. 73. Planes, which are perpendicular to the fame ftraight line, are parallel to one another.

Let the planes EF, GH, be perpendicular to the fame line AB; thefe planes are parallel. For, draw any ftraight line CD parallel to AB, meeting the planes in C and D, join AC, BD. Then CD as well as AB is perpendicular to both planes (Th. 49. Cor.); thus ABCD will be a rectangle, and confequently AB equal to CD, and in the fame way it may be fhewn, that all other perpendiculars terminated by both planes are equal; therefore the planes are parallel. (Def. 71.)

COR. Hence ftraight lines perpendicular to one of two parallel planes are alto perpendicular to the other plane.

THEOR. LII. fg. 76. If two ftraight lines be pa rallel to a third line, though not in the fame plane with it, they will be parallel to each other.

Let AB, CD, be each parallel to the fame line EF, though not in the fame plane with it, AB fhall be parallel to CD. For, let GH and GI be perpendicular to EF, in the planes AF and DE of the parallels; then fhall GF be perpendicular to the plane pathing by HGI (Tb. 48.); and HB, ID will alfo be perpendicular to the fame plane (Th. 49. Cor.), and therefore parallel. (Th. 49.)

THEOR. LIII. fig. 77. If two lines that meet each other, be parallel to two other lines that meet each other, tho' not in the same plane with them; the angles contained by thefe lines wi'l be equal.

Let the lines AB, AC, be parallel to the lines DE, DF, then will the angles BAC, EDF, be equal. For, take AB, AC, DE, DF, all equal, and join LB, FC, BC, EF. Then the lines AB, DE, being equal and parallel, the lines AD, BE, will allo be equal and parallel, (Th. 15.); and for the fame reafon AD, CF, are equal and parallei; therefore CF is parallel to BE, (Th. 52.) and alfo equal to it; hence BC is equal to EF. Thus the triangles ABC, DEF, are in all refpects equal, (Th. 5.); and therefore the angles BAC, EDF, are equal.

THEOR. LIV. fig. 78. The fections made by a plane cutting two parallel planes are also parallel to each other.

Let the parallel planes AB, CD, be cut by the 3d plane, EFG, in the lines EF, GH. Thefe lives are parallel. For, fuppofe EG, FH, to be drawn parallel to each other in the plane EFHG; alfo, let EI, FK, be perpendicular to the plane CD, and let IG, KH, be joined: Then EG, FH, being parallels, and EI, FK, being both perpendicular to the plane CD, are alfo parallel to each other, (Th. 49.) therefore the angle HFK is equal to the angle GEI (Th. 53.); but the angles FKH, EIG, are equal, being right angles; therefore the triangles FKH, EIG, are equiangular. (Th. 12, Cor. 3.), and the fides FK, EI, being equal, (Def. 71.), it follows, that the fides FH, EC, are also equal, (Th. 2.); but thefe two lines are parallel (by hyp.), as well as equal; therefore alfo EF and GH, which join their extremities are parallel. (Th. 15.)

We have now given the most material propofitions, with their demoftrations, of the elements of geometry, as far as relates to PLANE FIGURES, and to the pofitions and interfections of different planes. As to what relates to SOLID BODIES, fuch as the proportion of fimilar felids to one another, the proportion of Pyramids to Prisms, of the Cone to the Cylinder, and of the Sphere to the Cylin der, &c.; it can hardly be expected that in such a work as ours, we can find room for treating these parts of geometry in fo diffuse and rigid a manner, as they are treated of in books professedly written upon the fubject. We fhall therefore only recommend to fuch as wish to acquire the true spirit of geometrical reatoning, a caretul perufal of the works of Euclid and Archimedes; particularly the treati fes on his fphere and cylinder, and on conoids and fpheroids. In the 11th and 12th books of Euchd and in Archimedes's works, we may contemplate that very relined mode of geometrical reasoning,

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