PROP.XVI.—Theorem.
PROP.XVI.—Theorem.
If two parallel planes(AB,CD)be cut by a third plane(EF,HG), theircommon sections(EF,GH)with it are parallel.
Dem.—If the linesEF,GHare not parallel, they must meet at some finite distance. Let them meet inK. Now sinceKis a point in the lineEF, andEFis in the planeAB,Kis in the planeAB. In like mannerKis a point in the planeCD. Hence the planesAB,CDmeet inK, which is impossible, since they are parallel.Therefore the linesEF,GHmust beparallel.
Exercises.
Exercises.
1.Parallel planes intercept equal segments on parallel lines.
2.Parallel lines intersecting the same plane make equal angles with it.
3.A right line intersecting parallel planes makes equal angles with them.
PROP.XVII.—Theorem.
PROP.XVII.—Theorem.
If two parallel lines(AB,CD)be cut by three parallel planes(GH,KL,MN)intwo triads of points(A,E,B;C,F,D), their segments between those points areproportional.
Dem.—JoinAC,BD,AD. LetADmeet the planeKLinX. JoinEX,XF. Then because the parallel planesKL,MNare cut by the planeABDin the linesEX,BD, these lines are parallel [XI.xvi.]. Hence
AE : EB :: AX : XD [VI. ii.].
In like manner,
AX : XD :: CF : F D.
Therefore [V.xi.]
AE : EB :: CF : FD.
PROP.XVIII.—Theorem.
PROP.XVIII.—Theorem.
If a right line(AB)be normal to a plane(CI),anyplane(DE)passing through itshall be perpendicular to that plane.
Dem.—LetCEbe the common section of the planesDE,CI. From any pointFinCEdrawFGin the planeDEparallel toAB[I.xxxi.]. Then becauseABandFGare parallel, butABis normal, to the planeCI; henceFGis normal to it [XI.viii.]. Now sinceFGis parallel toAB, the anglesABF,BFGare equal to two right angles [I.xxix.]; butABFis right (hyp.); thereforeBFGis right—that is,FGis perpendicular toCE. Hence every line in the planeDE, drawn perpendicular to the common section of the planesDE,CI, is normal to the planeCI.Therefore[XI. Def.viii.]the planesDE,CIare perpendicular to eachother.
PROP.XIX.—Theorem.
If two intersecting planes(AB,BC)be each perpendicular to a third plane(ADC), their common section(BD)shall be normal to that plane.
Dem.—If not, draw fromDin the planeABthe lineDEperpendicular toAD, the common section of the planesAB,ADC; and in the planeBCdrawBFperpendicular to the common sectionDCof the planesBC,ADC. Then because the planeABis perpendicular toADC, the lineDEinABis normal to the planeADC[XI. Def.viii.]. In like mannerDFis normal to it. Therefore from the pointDthere are two distinct normals to the planeADC, which [XI.xiii.] is impossible.HenceBDmust be normal to the planeADC.
Exercises.
Exercises.
1.If three planes have a common line of intersection, the normals drawn to these planes fromany point of that line are coplanar.
2.If two intersecting planes be respectively perpendicular to two intersecting lines, the line ofintersection of the former is normal to the plane of the latter.
3.In the last case, show that the dihedral angle between the planes is equal to the rectilinealangle between the normals.
PROP.XX.—Theorem.The sum of any two plane angles(BAD,DAC)of a trihedral angle(A)isgreater than the third(BAC).
PROP.XX.—Theorem.The sum of any two plane angles(BAD,DAC)of a trihedral angle(A)isgreater than the third(BAC).
Dem.—If the third angleBACbe less than or equal to either of the other angles the proposition is evident. If not, suppose it greater: take any pointDinAD, and at the pointAin the planeBACmake the angleBAEequalBAD[I.xxiii.], and cut offAEequalAD. ThroughEdrawBC, cuttingAB,ACin the pointsB,C. JoinDB,DC.
Then the trianglesBAD,BAEhave the two sidesBA,ADin one equal respectively to the two sidesBA,AEin the other, and the angleBADequal toBAE; therefore the third sideBDis equal toBE. But the sum of the sidesBD,DCis greater thanBC; henceDCis greater thanEC. Again, because the trianglesDAC,EAChave the sidesDA,ACrespectively equal to the sidesEA,ACin the other, but the baseDCgreater thanEC; therefore [I.xxv.] the angleDACis greater thanEAC, but the angleDABis equal toBAE(const.).Hence the sum of the anglesBAD,DACis greater than the angleBAC.
PROP.XXI.—Theorem.
The sum of all the plane angles(BAC,CAD,&c.)forming any solid angle(A)isless than four right angles.
Dem.—Suppose for simplicity that the solid angleAis contained by five plane anglesBAC,CAD,DAE,EAF,FAB; and let the planes of these angles be cut by another plane in the linesBC,CD,DE,EF,FB; then we have [XI.xx.],
∠ABC+ABFgreater thanFBC,∠ACB+ACD,,BCD, &c.
∠ABC+ABFgreater thanFBC,∠ACB+ACD,,BCD, &c.
Hence, adding, we get the sum of the base angles of the five trianglesBAC,CAD, &c., greater than the sum of the interior angles of the pentagonBCDEF—that is, greater than six right angles. But the sum of the base angles of the same triangles, together with the sum of the plane anglesBAC,CAD, &c., forming the solid angleA, is equal to twice as many right angles as there are trianglesBAC,CAD, &c.—that is, equal to ten right angles.Hence the sum of the angles forming the solidangle is less than four right angles.
Observation.—This Prop.may not hold if the polygonal baseBCDEFcontainre-entrantangles.
Exercises on Book XI.
Exercises on Book XI.
1.Any face angle of a trihedral angle is less than the sum, but greater than the difference, of thesupplements of the other two face angles.
2.A solid angle cannot be formed of equal plane angles which are equal to the angles of aregular polygon ofnsides, except in the case ofn= 3,4,or5.
3.Through one of two non-coplanar lines draw a plane parallel to the other.
4.Draw a common perpendicular to two non-coplanar lines, and show that it is the shortestdistance between them.
5.If two of the plane angles of a tetrahedral angle be equal, the planes of these angles areequally inclined to the plane of the third angle, and conversely. If two of the planes of a trihedralangle be equally inclined to the third plane, the angles contained in those planes areequal.
6.The three lines of intersection of three planes are either parallel or concurrent.
7.If a trihedral angleObe formed by three right angles, andA,B,Cbe points along theedges, the orthocentre of the triangleABCis the foot of the normal fromOon the planeABC.
8.If through the vertexOof a trihedral angleO—ABCany lineODbe drawn interior to theangle, the sum of the rectilineal anglesDOA,DOB,DOCis less than the sum, but greater thanhalf the sum, of the face angles of the trihedral.
9.If on the edges of a trihedral angleO—ABCthree equal linesOA,OB,OCbetaken, each of these is greater than the radius of the circle described about the triangleABC.
10.Given the three angles of a trihedral angle, find, by a plane construction, the angles betweenthe containing planes.
11.If any planePcut the four sides of aGauchequadrilateral (a quadrilateral whose angularpoints are not coplanar)ABCDin four points,a,b,c,d, then the product of the fourratios
Aa- Bb Cc- Dd- aB , bC, cD, dA
is plus unity, and conversely, if the product
Aa Bb Cc Dd aB.bC-.cD-.dA-= +1,
the pointsa,b,c,dare coplanar.
12.If in the last exercise the intersecting plane be parallel to any two sides of the quadrilateral,it cuts the two remaining sides proportionally.
Def.x.—If at the vertexOof a trihedral angleO—ABCwe draw normalsOA′,OB′,OC′tothe facesOBC,OCA,OAB, respectively, in such a manner thatOA′will be on the same side of theplaneOBCasOA, &c., the trihedral angleO—A′B′C′is called the supplementaryof the trihedralangleO—ABC.
13.IfO—A′B′C′be the supplementary ofO—ABC, prove thatO—ABCis the supplementaryofO—A′B′C′.
14.If two trihedral angles be supplementary, each dihedral angle of one is the supplement of thecorresponding face angle of the other.
15.Through a given point draw a right line which will meet two non-coplanar lines.
16.Draw a right line parallel to a given line, which will meet two non-coplanar lines.
17.Being given an angleAOB, the locus of all the pointsPof space, such that the sum of theprojections of the lineOPonOAandOBmay be constant, is a plane.
APPENDIX.PRISM, PYRAMID, CYLINDER, SPHERE, AND CONE________________DEFINITIONS.
APPENDIX.PRISM, PYRAMID, CYLINDER, SPHERE, AND CONE________________DEFINITIONS.
i.Apolyhedronis a solid figure contained by plane figures: if it be contained by four plane figures it is called atetrahedron; by six, ahexahedron; by eight, anoctahedron; by twelve, adodecahedron; and if by twenty, anicosahedron.
ii.If the plane faces of a polyhedron be equal and similar rectilineal figures, it is called aregular polyhedron.
iii.Apyramidis a polyhedron of which all the faces but one meet in a point. This point is called thevertex; and the opposite face, thebase.
iv.Aprismis a polyhedron having a pair of parallel faces which are equal and similar rectilineal figures, and are called itsends. The others, called itsside faces, are parallelograms.
v.A prism whose ends are perpendicular to its sides is called arightprism; any other is called anobliqueprism.
vi.Thealtitudeof a pyramid is the length of the perpendicular drawn from its vertex to its base; and the altitude of a prism is the perpendicular distance between its ends.
vii.Aparallelopipedis a prism whose bases are parallelograms. A parallelopiped is evidently ahexahedron.
viii.Acubeis a rectangular parallelopiped, all whose sides are squares.
ix.Acylinderis a solid figure formed by the revolution of a rectangle about one of its sides, which remains fixed, and which is called itsaxis. The circles which terminate a cylinder are called itsbasesorends.
x.Aconeis the solid figure described by the revolution of a right-angled triangle about one of the legs, which remains fixed, and which is called theaxis. The other leg describes thebase, which is a circle.
xi.Asphereis the solid described by the revolution of a semicircle about a diameter, which remains fixed. Thecentreof the sphere is the centre of thegeneratingsemicircle. Any line passing through the centre of a sphere and terminated both ways by the surface is called adiameter.
PROP.I.—Theorem.
PROP.I.—Theorem.
Right prisms(ABCDE–FGHIJ,A′B′C′D′E′–F′G′H′I′J′)which have bases(ABCDE,A′B′C′D′E′)that are equal and similar, and which have equal altitudes,are equal.
Dem.—Apply the bases to each other; then, since they are equal and similar figures, they will coincide—that is, the pointAwithA′,BwithB′, &c. And sinceAFis equal toA′F′, and each is normal to its respective base, the: pointFwill coincide withF′. In the same manner the pointsG,H,I,Jwill coincide respectively with the pointsG′,H′,I′,J′.Hence the prisms are equal in everyrespect.
Cor. 1.—Right prisms which have equal bases(EF,E′F′)and equal altitudes areequal in volume.
Dem.—Since the bases are equal, but not similar, we can suppose one of them,EF, divided into partsA,B,C, and re-arranged so as to make them coincide with the other [I.xxxv., note]; and since the prism onE′F′can be subdivided in the same manner by planes perpendicular to the base, the proposition is evident.
Cor. 2.—The volumes of right prisms(X,Y)having equal bases are proportionalto their altitudes.
For, if the altitudes be in the ratio ofm:n,Xcan be divided intomprisms of equal altitudes by planes parallel to the base; then thesemprisms will be all equal. In like manner,Ycan be divided intonequal prisms.HenceX:Y::m:n.
Cor. 3.—In right prisms of equal altitudes the volumes are to one another as the areas of their bases. This may be proved by dividing the bases into parts so that the subdivisions will be equal, and then the volumes proportional to the number of subdivisions in their respective bases, that is, to their areas.
Cor. 4.—The volume of a rectangular parallelopiped is measured by the continued product of its three dimensions.
PROP.II.—Theorem.Parallelopipeds(ABCD–EFGH,ABCD–MNOP), having a common base(ABCD)and equal altitudes, are equal.
PROP.II.—Theorem.Parallelopipeds(ABCD–EFGH,ABCD–MNOP), having a common base(ABCD)and equal altitudes, are equal.
1∘. Let the edgesMN,EFbe in one right line; thenGH,OPmust be in one right line. NowEF=MN, because each equalAB; thereforeME=NF; therefore the prismsAEM–DHP, andBFN–CGO, have their triangular basesAEM,BFNidentically equal, and they have equal altitudes; hence they are equal; and supposing them taken away from the entire solid,the remaining parallelopipedsABCD–EFGH,ABCD–MNOPare equal.
2∘. Let the edgesEF,MNbe in different lines; then produceON,PMto meet the linesEFandGHproduced in the pointsJ,K,L,I. Then by 1∘the parallelopipedsABCD–EFGH,ABCD–MNOPare each equal to the parallelopipedABCD–IJKL.Hence they an equal to one another.
Cor.—The volume of any parallelopiped is equal to the product of its base and altitude.
PROP.III.—Theorem.A diagonal plane of a parallelopiped divides it into two prisms of equal volume.
PROP.III.—Theorem.A diagonal plane of a parallelopiped divides it into two prisms of equal volume.
1∘. When the parallelopiped is rectangular the proposition is evident.
2∘. When it is any parallelopiped,ABCD–EFGH, the diagonal plane bisects it.
Dem.—Through the verticesA,Elet planes be drawn perpendicular to the edges and cutting them in the pointsI,J,K;L,M,N, respectively. Then [I.xxxiv.] we haveIL=BF, because each is equal toAE. HenceIB=LF. In like mannerJC=MG. Hence the pyramidA–IJCBagrees in everything but position withE–LMGF; hence it is equal to it in volume. To each add the solidABC–LME, and we have the prismAIJ–ELMequal to the prismABC–EFG. In like mannerAJK–EMN=ACD–EGH; but (1∘) the prismAIJ–ELM=AJK–EMN. HenceABC–EFG=ACD–EGH.Therefore the diagonal plane bisects theparallelopiped.
Cor.1.—The volume of a triangular prism is equal to the product of its base and altitude; because it is half of a parallelopiped, which has a double base and equal altitude.
Cor.2.—The volume of any prism is equal to the product of its base and altitude; because it can be divided into triangular prisms.
PROP.IV.—Theorem.If a pyramid(O–ABCDE)be cut by any plane(abcde)parallel to the base, thesection is similar to the base.
PROP.IV.—Theorem.If a pyramid(O–ABCDE)be cut by any plane(abcde)parallel to the base, thesection is similar to the base.