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E+2 ED. But (Theorem XXXV.) BD+ BC=2 DE2+ 2 B F2, and consequently 2 D E2 + 2 BE2 = 2 A B2 + 2 A E2 + 2 E D2; or, by omitting 2 D E from each, 2 BE 2 A B' +2A E'; or B E2 = A B2 + A E2; and therefore the angle BAE is a right angle. AB is therefore at right angles to any line which it meets on the plane of AC, and AD, and it is consequently at right angles to that plane.

THEOREM LXV.—If a straight line as AB (fig. 85, p. II.) be at right angles to each of three straight lines, as BC, BD, BE, at B, their point of meeting, then these straight lines are all in one plane.

If they are not, let the plane passing through BE and BD meet the plane passing through BA and BC in the line BF; then BA (The orem LXIV.) being perpendicular to the plane, BDE is perpendicular to BF, which it meets in that plane; hence the angles ABC and ABF in the same plane are equal, the less to the greater, which is impossible. Hence BC cannot be out of the plane in which BE and ED are, and the three lines are consequently in the same plane. THEOREM LXVI. If two straight lines as AB, CD (fig. 86, p. II.), be at right angles to the same plane, as BD E, they are parallel to each other.

In the plane B D E draw DE at right angles to BD; take any point E in it, and join E A, EB, and AD. Then (Theorem LIV. Cor. 2) A E2 =AB2 + BE2 = A B2 + B D2 + DEAD' +DE; therefore the angle ADC is a right angle; and, as the angles BDC and CDA are right angles, the lines CD, and A D, and BD, are in the same plane (Theorem LXIV.). But (Theorem LXII.) AB is in the plane of AD and BD; hence AB and CD are in the same plane, and, as the angles A B D and CDB are right angles, the lines A B and CD are parallel. THEOREM LXVII.-If A B and C D (fig. 87,) p. II.) are parallel, and one of them, as A B, is perpendicular to a plane, as E F, the other, CD. is perpendicular to the same plane.

For if CD is not perpendicular to the plane EF, let DC, drawn through G, be perpendicular to A. Then (Theorem LXVI.) A B and DG are parallel; consequently DC and DG which cut each other, are both parallel to CD, which is impossible. Therefore no line drawn through D, except DC, is perpendicular to the plane EF; and DC is consequently perpendicular to A.

THEOREM LXVIII.-If two lines, as AB, CD (fig. 88, p. II.), be each parallel to another line, as EF, though not in the same plane with it, these two lines are parallel to each other. From any point G, in EF, draw G H, in the plane of AB and F F, and perpendicular to EF; and from the same point G, but in the plane of CD and EF, draw GK perpendicular to EF. Then (Theorem LXIV.) EG is perpendicular to the plane KGH, and therefore (Theorem LXVII.) A B and CO are both perpendicular to the same plane; and consequently (Theorem LXVI.) they are parallel to each other.

THEOREM LXIX.-If two straight lines meeting each other, as A B, BC, (fig. 89, p. II.), are parallel to two other straight lines, as DE, EF,

which also meet each other, but are not in the same plane with A B and B C, the angles A B C and D E F are equal.

For take A B equal to DE, BC to EC, and join A C, DF, AD, BE, and CF. Then A B being equal and parallel to DE, AD and BE are equal and parallel (Theorem XXIX.), and for the latter reason BE and CF are parallel. Hence AD and CF are equal and parallel, and therefore A C and D F are equal and parallel; consequently (Theorem V.) the angles A B C and D E F are equal.

THEOREM LXX.-If two planes as AB, C D (fig. 90, p. III.) be cut by a third plane, EH, the sections E F and GH are parallel.

For in the plane EH let EF and G H be drawn parallel to each other, and let EI and FK be perpendicular from E and F on the plane CD; then (Theorem LXVI.) EL and FK are parallels; and consequently (Theorem LXIX.) the angles HFK and GEI are equal, and the right angles EIG and FHK being equal, as well as F K and EL, the distance of the parallel planes; the triangles FHK and EGI are identical, and have FH equal to EG; but their lines are also parallel, therefore the lines EF and HG, which join them, are parallel (Theorem XXIII.)

THEOREM LXXI.-If a straight line, as AB (fig. 91, p. III.), be perpendicular to a plane, as CK, any plane DE, passing through AB, is at right angles to the plane CK.

For from any point F, in CE, the common section of the two planes, draw FG, in the plane DE, perpendicular to C E. Then AB, being perpendicular to C K, is perpendicular to CE, which it meets in that plane, and it is therefore parallel to FG; and consequently (Theorem LXVIII.) FG is perpendicular to the plane CK. Hence (Def. 3, PLANES), the plane CH is perpendicular to the plane C K.

THEOREM LXXII.-If each of two planes, as A B, BC (fig. 92. p. III.), be perpendicular to another plane, as ADC, the common section, B D, of the first two planes, is perpendicular to the third plane.

For from D, in the plane ADC, draw DE perpendicular to AD, and DF to DC. Then as DE is perpendicular to DA, the common section of the planes AB and A DC, and those planes are at right angles to each other, ED is at right angles to AB, and consequently at right angles to D B, which it meets in that plane. For the same reason, DF is at right angles to DB; and hence, as BD is at right angles to DF and DE, it is at right angles to the plane ADC, in which those lines are (Theorem LXIV.)

THEOREM LXXII-If a solid angle, as A (fig. 93, p. III.) be contained by three plane angles as CAB, CAD, DA B, any two of these angles together are greater than the third.

If the angles are all equal, the proposition is manifest.-If they are not equal, let BAC be the greatest, and, in the plane BA C, draw A E. making the angle BAE equal to the angle BAD; make A E equal to AD, and through E draw any straight line BEC, cutting A B and AC in B and C, and join BD, CD. Then (Theorem I.) the triangles BA E and BAD are

identical, and have BD and B E equal, and consequently EC is the difference of B C and BD, and (Theorem XIV.) EC is less than DC. Now as AE is equal to AD, AC common to both the triangles ACE and ACD, but EC less than DC, it follows (Theorem XII.) that the angle EAC is less than the angle CAD. Hence the angles BAE and EAC together, or the whole angle BAC, is less than the sum of the angles BAD and CAD.

THEOREM LXXIV.-If a solid angle as A (fig. 94, p. III.) be contained by any number of plane angles, as BAC, CAD, DAB, EAB, these plane angles together are less than four right angles.

Let the planes which contain the solid angle, A, be cut by another plane BCDE. Then the solid angles at B, C, D, and E, being each contained by three plane angles, A BE and ABC are less than CBE, AC B, ACD are less than BCD, ADC and AD E are less than CDE, and A E B and A E D are less than BED. Hence the angles at the bases of the triangles which have their common vertex at A are greater than the interior angles of the rectilineal figure BCD E. But all the angles of the triangles BAC, CAD, DAE and BA E, are equal to twice as many angles as there are sides in the figure BCDE; and the interior angles of that figure, together with four right angles, being also equal to twice as many right angles as the figure has sides (Theorem XX.), the angles of the triangles are equal to the interior angles of the plane figure and four right angles. And, as all the angles at the bases of the triangles are greater than all the interior angles of the figure; the remaining angles of the triangles, or those which contain the solid angle A, are less than four right angles.

ON THE COMPARISON OF SOLIDS.

DEFINITIONS.

1. Similar solids, contained by plane figures hare their corresponding solid angles equal, and are bounded by the same number of similar planes, alike placed.

2. A prism is a solid 'whose ends are parallel, equal, and similar plane figures, and its sides, connecting those ends, are parallelograms.

3. An upright prism is one having the planes of its sides perpendicular to the planes of its ends.

4. A parallelopiped, or parallelopipedon, is a prism bounded by six parallelograms, every opposite two of which are equal, similar, and parallel.

5. A rectangular parallelopipedon is one whose bounding planes are rectangles, perpendicular to each other.

6. A cube is a rectangular parallelopipedon, whose sides are all equal.

7. A cylinder is a prism whose ends are circles; and its axis is the right line joining the centres of the parallel circles which form its ends. 8. A pyramid is a solid whose base is any rectilineal plane figure, and its sides triangles whose vertices meet in a point above the base, called the vertex of the pyramid.

9. A cone is a pyramid having a circle for its

base; and the axis of a cone is the line joining the vertex to the centre of the circle which forms the base.

10. Similar cones, and cylinders, are those that have their altitudes in the same proportions as the diameters of their bases.

11. A sphere is a solid bounded by a curve surface, every point of which is at the same distance from a point within, called the centre.

12. The diameter of a sphere is any right line passing through the centre and terminated both ways by the surface.

13. The altitude of a solid is the perpendicular drawn from the vertex to the opposite side or base.

THEOREM LXXV.-A section of any pyramia as ABCD (fig. 95, p. III.), parallel to the base BCDE, is similar to the base; and these two planes are to each other as the squares of their distance from the vertex, or as AH to A I3, AIH being a perpendicular from A on the two parallel planes.

For join CH, FI. Then (Theorem LXX.) BC and E F are parallel, and C D and FG are parallel, and consequently (Theorem LXVIII.) the angles E F G and BCD are equal. In the same way it may be shown that each angle in the plane E G is equal to the corresponding angle in the plane BD, and consequently these planes are equiangular.

Again the triangles ABC and AEF being equiangular, as are also the triangles ACD and AF G, we have (Theorem LII. Čor.) AC: AF ::BC: EF:: CD: FG. In the same way it may be shown that all the sides of the plane of ÉG are proportional to the corresponding ones in BD, hence (Def. 52) the figures BCDE and EFGO are similar.

But (Theorem LXI.) the plane BD is to the plane E G as BC: EF', or as A C2: A F2; or as A H3: A 12; because the triangles AHC, AIF, having the angles H and I right angles, and the angle A common, are similar, and therefore A C :AF:: AH: AI, or A C2: A F2 :: A H2: A I2, whence the plane BD is to the plane EG as A H2: A I2.

Cor.-If the point A be conceived to be infinitely distant from the base, the pyramid will become a prism; and the ratio of AH to A I, as well as that of A H2 to A I3, will be that of equality. Hence in any prism a section by a plane parallel to the base will be equal and similar to the base.

THEOREM LXXVI.-In any cone as ABCD (fig. 96, p. III.), if GHI be a section parallel to the base, then G H I is a circle; and BCD, GHI are to each other as the squares of their distances from the vertex.

For let A LF be perpendicular to the parallel planes, and let the planes ACE, A DE, pass through the axis A KE of the cone, meeting the section in the points H, I, K.

Then (Theorem LXX.) H K and EC are parallel, as well as KI and ED, and by similar triangles K II: EC::AK:AE::KL: ED; but EC is equal to E D, therefore K I is equal to K H; and, as the same may be shown of any other lines drawn from K to the circumference of the section G H I, that section is a circle.

Again, by similar triangles AL: AF:: AK :AE: KI: ED, whence AL': A F2:: KI: E D2:: circle G H I: circle DCD (The orem LXI. Cor). Hence A L2: A F2:: circle GHI: circle BC D.

Cor. If the point A be conceived to be infinitely distant from the base, the cone will become a cylinder, and the ratio of AL to A F, as well as that of AL' to A F3, will be that of equality. Hence in any cylinder, the sections parallel to the base are circles equal to the base.

THEOREM LXXVII.—All pyramids and cones of equal bases and altitudes are equal to each other.

Let the pyramid ABCDI (fig. 95, p. III), and the cone ABCD (fig. 96, p. III.) have equal bases and altitudes, and parallel to these bases, and at equal distances, AI and AL, conceive the planes EG and GKHI to be drawn. Then (Theorem LXXV. and LXXVI.) AI: A H3 :: EG: BD; and A L2: A Fa: G KHI; BCD; and as A I3, A H3, are equal to A L3, AF, therefore EG: BD::GKHI BCD; and, as BD is equal to BCD, EG is equal to GKHI. In this manner may all rations in each figure at the same distance from the vertex be shown to be equal, and consequently, as the heights are equal, the solids which are composed of these sections are equal.

Cor. All prisms and cylinders whose bases and altitudes are equal, are equal to each other; and prisms and cylinders are equal to a rectangular parallelopipedon of the same base and altitude.

THEOREM LXXVIII.-A pyramid whose base 13 triangular, as BDEF (fig. 97, p. III.), is the third part of a prism having the same base and altitude.

For let ABCDEF be a prism on the same triangular base DEF, and, on the three rectangular sides of the prism, draw the diagonals BF, BD, CD. Then the planes BDF, B.CD, divide the whole prism into three pyramids, BDEF, DABC, DBCF, which may be thus proved equal to each other.

The bases ABC and DEF being equal (Def. 2, SOLIDS), the pyramids ABCD and DEFB are equal (Theorem LXXVII.); and the bases BEF and BCF being equal (Theorem XXI.) the pyramids DEFB and BCFD are equal (Theorem LXX.). Hence all the three pyramids which compose the prism are equal, and the pyramid is the third part of the prism, or the prism is three times the pyramid.

Cor. 1. Every pyramid, whatever may be its figure, is the third part of a prism having the same base and altitude, since the base of the prisin may be divided into triangles, and the whole solid into triangular prisms and pyramids. Cor. 2. A cone is the third part of a cylinder, or of a prism having the same base and altitude; for it has been proved that a cylinder is equal to a prism, and a cone equal to a pyramid of equal base and altitude.

THEOREM LXXIX.-If AC and EG (fig. 98, p. III.) be two rectangular parallelopipedons having equal altitudes, AD, FH, then AC is to EG as the base AB to the base EF.

For, let the base AB be to the base EF as any number (say 3) is to any other number

(as 2); and conceive AB to be divided into three equal rectangles, AI, LK, and MB; and EF into two equal rectangles, EO and PF; and through the lines of division let the planes LR, MS, PV, pass, parallel to A Q, and ET. Then the parallelopipedons AR, LS, MC, EV, and PQ, having equal bases and altitudes, are all equal (Theorein LXXVII). Hence the solids are to each other in the same proportion as their bases. Cor. 1. From this theorem, (and Theorem LXXVII. Cor.) it appears that all prisms and cylinders of equal altitudes are to each other as their bases; and from this, combined with Theorem XLV. Cor., and Theorem LXXVIII., it appears that pyramids and cones of equal altitudes are to each other as their bases.

Cor. 2. By considering AQ and TE as the equal bases of the parallelopipedon AC, EG, those parallelopipedons will be to each other as their altitudes AN, EW. Hence prisms and cylinders of equal bases are to each other as their altitudes, and consequently pyramids and cones of equal bases are also to each other as their altitudes.

Cor. 3. As prisms and cylinders are as their altitudes when their bases are equal, and as their bases when their altitudes are equal, therefore, when neither are equal, they are in the compound ratio of their bases and altitudes.

THEOREM LXXX.-Similar prisms and cylinders are to each other as the cubes of their like linear dimensions.

For the bases are as the squares of their like sides, and the altitudes are as those sides; and the solidities being in the compound ratio of the bases and altitudes (Theorem LXXIX., Cor. 2) are as the cubes of those like sides.

Cor. Similar pyramids and similar cones, being the third parts of their corresponding prisms and cylinders, are to each other as the cubes of their like linear dimensions; and all similar solids whatever, being composed of similar pyramids, are to each other as the cubes of their like linear dimensions.

THEOREM LXXXI.-Every section of a sphere by a plane, as CDEGF (fig. 99, p. III.), is a circle.

If the plane pass through the centre, then, as every point in the surface of the sphere is equidistant from its centre, the section is a plane figure, every point of whose periphery is equidistant from a certain point within it, and the figure is therefore a circle.

But if the plane do not pass through the ceutre, from the centre A, let AB be a perpendicular to the plane; take any two points, C, D, in the circumference of the section, and join AC, AD, BC, and BD. Then, as AB is perpendicular to the plane CDEGF, it is perpendicular to the lines CB and B D, which it meets in that plane; therefore ACA B+ B C2, and A DAB+ BD; hence, as ACAD, A B+ BC AB2 + BD, and consequently BC BD', or BC BD. Hence all lines drawn from B in the plane CDEGF, to the periphery of that plane, are equal, and the figure is therefore a circle.

=

THEOREM LXXXII.-A sphere is two-thirds of its circumscribing cylinder.

For, let AC (fig. 100, p. III.) be a section of

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