PROP.XXXV.—Theorem.Parallelograms on the same base(BC)and between the same parallels areequal.
PROP.XXXV.—Theorem.Parallelograms on the same base(BC)and between the same parallels areequal.
Dem.—1. Let the sidesAD,DFof the parallelogramsAC,BFopposite to the common baseBCterminate in the same pointD, then [xxxiv.] each parallelogram is double of the triangleBCD.Hence they are equal to oneanother.
2. Let the sidesAD,EF(figures (α), (β)) opposite toBCnot terminate in the same point.
Then becauseABCDis a parallelogram,ADis equal toBC[xxxiv.]; and sinceBCEFis a parallelogram,EFis equal toBC; therefore (see fig. (α)) take awayED, and in fig. (β) addED, and we have in each caseAEequal toDF, andBAis equal toCD[xxxiv.]. Hence the trianglesBAE,CDFhave the two sidesBA,AEin one respectively equal to the two sidesCD,DFin the other, and the angleBAE[xxix.] equal to the angleCDF; hence [iv.] the triangleBAEis equal to the triangleCDF; and taking each of these triangles in succession from the quadrilateralBAFC,there will remain the parallelogramBCFEequal to the parallelogramBCDA.
Or thus:The trianglesABE,DCFhave [xxxiv.] the sidesAB,BEin one respectively equal to the sidesDC,CFin the other, and the angleABEequal to the angleDCF[xxix., Ex. 8]. Hence the triangleABEis equal to the triangleDCF; and, taking each away from the quadrilateralBAFC,there will remain the parallelogramBCFEequal to the parallelogramBCDA.
Observation.—By the second method of proof the subdivision of the demonstration into cases isavoided. It is easy to see that either of the two parallelogramsABCD,EBCFcan be divided intoparts and rearranged so as to make it congruent with the other. This Proposition affords thefirst instance in the Elements in which equality which is not congruence occurs. Thisequality is expressed algebraically by the symbol =, while congruence is denoted by≡,called also the symbol of identity. Figures that are congruent are said to beidenticallyequal.
PROP.XXXVI.—Theorem.Parallelograms(BD,FH)on equal bases(BC,FG)and between the sameparallels are equal.
PROP.XXXVI.—Theorem.Parallelograms(BD,FH)on equal bases(BC,FG)and between the sameparallels are equal.
Dem.—JoinBE,CH. Now sinceFHis a parallelogram,FGis equal toEH[xxxiv.]; butBCis equal toFG(hyp.); thereforeBCis equal toEH(Axiomi.). HenceBE,CH, which join their adjacent extremities, are equal and parallel; thereforeBHis a parallelogram. Again, since the parallelogramsBD,BHare on the same baseBC, and between the same parallelsBC,AH, they are equal [xxxv.]. In like manner, since the parallelogramsHB,HFare on the same baseEH, and between the same parallelsEH,BG, they are equal. HenceBDandFHare each equal toBH.Therefore(Axiomi.)BDis equal toFH.
Exercise.—Prove this Proposition without joiningBE,CH.
PROP.XXXVII.—Theorem.Triangles(ABC,DBC)on the same base(BC)and between the sameparallels(AD,BC)are equal.
PROP.XXXVII.—Theorem.Triangles(ABC,DBC)on the same base(BC)and between the sameparallels(AD,BC)are equal.
Dem.—ProduceADboth ways. DrawBEparallel toAC, andCFparallel toBD[xxxi.] Then the figuresAEBC,DBCFare parallelograms; and since they are on the same baseBC, and between the same parallelsBC,EFthey are equal [xxxv.]. Again, the triangleABCis half the parallelogramAEBC[xxxiv.], because the diagonalABbisects it. In like manner the triangleDBCis half the parallelogramDBCF, because the diagonal DC bisects it, and halves of equal things are equal (Axiomvii.).Therefore the triangleABCis equal to the triangleDBC.
Exercises.
Exercises.
1.If two equal triangles be on the same base, but on opposite sides, the right line joining theirvertices is bisected by the base.
2.Construct a triangle equal in area to a given quadrilateral figure.
3.Construct a triangle equal in area to a given rectilineal figure.
4.Construct a lozenge equal to a given parallelogram, and having a given side of theparallelogram for base.
5.Given the base and the area of a triangle, find the locus of the vertex.
6.If through a pointO, in the production of the diagonalACof a parallelogramABCD, anyright line be drawn cutting the sidesAB,BCin the pointsE,F, andED,FDbe joined, thetriangleEFDis less than half the parallelogram.
PROP.XXXVIII.—Theorem.Two triangles on equal bases and between the same parallels are equal.
PROP.XXXVIII.—Theorem.Two triangles on equal bases and between the same parallels are equal.
Dem.—By a construction similar to the last, we see that the triangles are the halves of parallelograms, on equal bases, and between the same parallels. Hence they are the halves of equal parallelograms [xxxvi.].Therefore they are equal to oneanother.
Exercises.
Exercises.
1.Every median of a triangle bisects the triangle.
2.If two triangles have two sides of one respectively equal to two sides of the other, and thecontained angles supplemental, their areas are equal.
3.If the base of a triangle be divided into any number of equal parts, right lines drawn fromthe vertex to the points of division will divide the whole triangle into as many equalparts.
4.Right lines from any point in the diagonal of a parallelogram to the angular points throughwhich the diagonal does not pass, and the diagonal, divide the parallelogram into four triangleswhich are equal, two by two.
5.If one diagonal of a quadrilateral bisects the other, it also bisects the quadrilateral, andconversely.
6.If two△sABC,ABDbe on the same baseAB, and between the same parallels, and if aparallel toABmeet the sidesAC,BCin the pointE,F; and the sidesAD,BDin the pointG,H;thenEF=GH.
7.If instead of triangles on the same base we have triangles on equal bases and between thesame parallels, the intercepts made by the sides of the triangles on any parallel to the bases areequal.
8.If the middle points of any two sides of a triangle be joined, the triangle so formed with thetwo half sides is one-fourth of the whole.
9.The triangle whose vertices are the middle points of two sides, and any point in the base ofanother triangle, is one-fourth of that triangle.
10.Bisect a given triangle by a right line drawn from a given point in one of thesides.
11.Trisect a given triangle by three right lines drawn from a given point withinit.
12.Prove that any right line through the intersection of the diagonals of a parallelogram bisectsthe parallelogram.
13.The triangle formed by joining the middle point of one of the non-parallel sides of atrapezium to the extremities of the opposite side is equal to half the trapezium.
PROP.XXXIX.—Theorem.Equal triangles(BAC,BDC)on the same base(BC)and on the same sideof it are between the same parallels.
PROP.XXXIX.—Theorem.Equal triangles(BAC,BDC)on the same base(BC)and on the same sideof it are between the same parallels.
Dem.—JoinAD. Then ifADbe not parallel toBC, letAEbe parallel to it, and let it cutBDinE. JoinEC. Now since the trianglesBEC,BACare on the same baseBC, and between the same parallelsBC,AE, they are equal [xxxvii.]; but the triangleBACis equal to the triangleBDC(hyp.). Therefore (Axiomi.) the triangleBECis equal to the triangleBDC—that is, a part equal to the whole which is absurd.HenceADmust be parallel toBC.
PROP.XL.—Theorem.
PROP.XL.—Theorem.
Equal triangles(ABC,DEF)on equal bases(BC,EF)which form parts ofthe same right line, and on the same side of the line, are between the sameparallels.
Dem.—JoinAD. IfADbe not parallel toBF, letAGbe parallel to it. JoinGF. Now since the trianglesGEFandABCare on equal basesBC,EF, and between the same parallelsBF,AG, they are equal [xxxviii.]; but the triangleDEFis equal to the triangleABC(hyp.). HenceGEFis equal toDEF(Axiomi.)—that is, a part equal to the whole, which is absurd.ThereforeADmust be parallel toBF.
Def.—The altitude of a triangle is the perpendicular from the vertex on thebase.
Exercises.
Exercises.
1.Triangles and parallelograms of equal bases and altitudes are respectively equal.
2.The right line joining the middle points of two sides of a triangle is parallel to the third; forthe medians from the extremities of the base to these points will each bisect the original triangle.Hence the two triangles whose base is the third side and whose vertices are the points of bisectionare equal.
3.The parallel to any side of a triangle through the middle point of another bisects thethird.
4.The lines of connexion of the middle points of the sides of a triangle divide it into fourcongruent triangles.
5.The line of connexion of the middle points of two sides of a triangle is equal to half the thirdside.
6.The middle points of the four sides of a convex quadrilateral, taken in order, are the angularpoints of a parallelogram whose area is equal to half the area of the quadrilateral.
7.The sum of the two parallel sides of a trapezium is double the line joining the middle pointsof the two remaining sides.
8.The parallelogram formed by the line of connexion of the middle points of two sides of atriangle, and any pair of parallels drawn through the same points to meet the third side, is equal tohalf the triangle.
9.The right line joining the middle points of opposite sides of a quadrilateral, and the right linejoining the middle points of its diagonals, are concurrent.
PROP.XLI.—Theorem.
PROP.XLI.—Theorem.
If a parallelogram(ABCD)and a triangle(EBC)be on the same base(BC)andbetween the same parallels, the parallelogram is double of the triangle.
Dem.—JoinAC. The parallelogramABCDis double of the triangleABC[xxxiv.]; but the triangleABCis equal to the triangleEBC[xxxvii.].Therefore theparallelogramABCDis double of the triangleEBC.
Cor.1.—If a triangle and a parallelogram have equal altitudes, and if the base of the triangle be double of the base of the parallelogram, the areas are equal.
Cor.2.—The sum of the triangles whose bases are two opposite sides of a parallelogram, and which have any point between these sides as a common vertex, is equal to half the parallelogram.
PROP.XLII.—Problem.To construct a parallelogram equal to a given triangle(ABC), and having anangle equal to a given angle(D).
PROP.XLII.—Problem.To construct a parallelogram equal to a given triangle(ABC), and having anangle equal to a given angle(D).
Sol.—BisectABinE. JoinEC. Make the angleBEF[xxiii.] equal toD. DrawCGparallel toAB[xxxi.], andBGparallel toEF.EGisa parallelogram fulfillingthe required conditions.
Dem.—BecauseAEis equal toEB(const.), the triangleAECis equal to the triangleEBC[xxxviii.], therefore the triangleABCis double of the triangleEBC; but the parallelogramEGis also double of the triangleEBC[xli.], because they are on the same baseEB, and between the same parallelsEBandCG. Therefore the parallelogramEGis equal to the triangleABC, and it has (const.) the angleBEFequal toD.HenceEGis a parallelogram fulfilling the requiredconditions.
PROP.XLIII.—Theorem.
PROP.XLIII.—Theorem.
The parallels(EF,GH)through any point(K)in one of the diagonals(AC)of aparallelogram divide it into four parallelograms, of which the two(BK,KD)throughwhich the diagonal does not pass, and which are called thecomplementsof theother two, are equal.
Dem.—Because the diagonal bisects the parallelogramsAC,AK,KCwe have [xxxiv.] the triangleADCequal to the triangleABC, the triangleAHKequal toAEK, and the triangleKFCequal to the triangleKGC. Hence, subtracting the sums of the two last equalities from the first, we getthe parallelogramDKequal tothe parallelogramKB.
Cor.1.—If through a pointKwithin a parallelogramABCDlines drawn parallel to the sides make the parallelogramsDK,KBequal,Kis a point in the diagonalAC.
Cor.2.—The parallelogramBHis equal toAF, andBFtoHC.
Cor.2. supplies an easy demonstration of a fundamental Proposition in Statics.
Exercises.
Exercises.
1.IfEF,GHbe parallels to the adjacent sides of a parallelogramABCD, the diagonalsEH,GFof two of the fourPICTs into which they divide it and one of the diagonals ofABCDareconcurrent.
Dem.—LetEH,GFmeet inM; throughMdrawMP,MJparallel toAB,BC. ProduceAD,GH,BCto meetMP, andAB,EF,DCto meetMJ. Now the complementOF=FJ: to each addthePICTFL, and we get the figureOFL=PICTCJ. Again, the complementPH=HK[xliii.]: to each add thePICTOC, and we get thePICTPC=figureOFL. Hence thePICTPC=CJ. Therefore they are about the same diagonal [xliii.,Cor.1].HenceACproduced will pass throughM.
2.The middle points of the three diagonalsAC,BD,EFof a quadrilateralABCDarecollinear.
Dem.—Complete thePICTAEBG. DrawDH,CIparallel toAG,BG. JoinIH, andproduce; thenAB,CD,IHare concurrent (Ex.1); thereforeIHwill pass throughF. JoinEI,EH.Now [xi., Ex.2, 3] the middle points ofEI,EH,EFare collinear, but [xxxiv., Ex.1] the middlepoints ofEI,EHare the middle points ofAC,BD.Hence the middle points ofAC,BD,EFarecollinear.
PROP.XLIV.—Problem.
PROP.XLIV.—Problem.
To a given, right line(AB)to apply a parallelogram which shall be equalto a given triangle(C), and have one of its angles equal to a given angle(D).
Sol.—Construct the parallelogramBEFG[xlii.] equal to the given triangleC, and having the angleBequal to the given angleD, and so that its sideBEshall be in the same right line withAB. ThroughAdrawAHparallel toBG[xxxi.], and produceFGto meet it inH. JoinHB. Then becauseHAandFEare parallels, andHFintersects them, the sum of the anglesAHF,HFEis two right angles [xxix.]; therefore the sum of the anglesBHF,HFEis less than two right angles; and therefore (Axiomxii.) the linesHB,FE, if produced, will meet as atK. ThroughKdrawKLparallel toAB[xxxi.], and produceHAandGBto meet it in the pointsLandM.ThenAMis a parallelogram fulfilling the requiredconditions.
Dem.—The parallelogramAMis equal toGE[xliii.]; butGEis equal to the triangleC(const.); thereforeAMis equal to the triangleC. Again, the angleABMis equal toEBG[xv.], andEBGis equal toD(const.); therefore the angleABMis equal toD; andAMis constructed on the given line;therefore it is the parallelogramrequired.
PROP.XLV.—Problem.To construct a parallelogram equal to a given rectilineal figure(ABCD), andhaving an angle equal to a given rectilineal angle(X).
PROP.XLV.—Problem.To construct a parallelogram equal to a given rectilineal figure(ABCD), andhaving an angle equal to a given rectilineal angle(X).