Exercises.
Exercises.
1.Triangles which have one angle of one equal or supplemental to one angle of the other, are toone another in the ratio of the rectangles of the sides about those angles.
2.Two quadrilaterals whose diagonals intersect at equal angles are to one another in the ratio ofthe rectangles of the diagonals.
PROP.XXIV.—Theorem.In any parallelogram(AC), every two parallelograms(AF,FC)which areabout a diagonal are similar to the whole and to one another.
PROP.XXIV.—Theorem.In any parallelogram(AC), every two parallelograms(AF,FC)which areabout a diagonal are similar to the whole and to one another.
Dem.—Since the parallelogramsAC,AFhave a common angle, they are equiangular [I.xxxiv.], and all that is required to be proved is, that the sides about the equal angles are proportional. Now, since the linesEF,BCare parallel, the trianglesAEF,ABCare equiangular; therefore [iv.]AE:EF::AB:BC, and the other sides of the parallelograms are equal toAE,EF;AB,BC: hence the sides about the equal angles are proportional;therefore the parallelogramsAF,ACare similar.In the same mannerthe parallelogramsAF,FCaresimilar.
Cor.—The parallelogramsAF,FC,ACare, two by two, homothetic.
PROP.XXV.—Problem.To describe a rectilineal figure equal to a given one(A), and similar to anothergiven one(BCD).
PROP.XXV.—Problem.To describe a rectilineal figure equal to a given one(A), and similar to anothergiven one(BCD).
Sol.—On any sideBCof the figureBCDdescribe the rectangleBEequal toBCD[I.xlv.], and onCEdescribe the rectangleEFequal toA. BetweenBC,CFfind a mean proportionalGH, and on it describe the figureGHIsimilar toBCD[xviii.], so thatBCandGHmay be homologous sides.GHIis the figurerequired.
Dem.—The three linesBC,GH,CFare in continued proportion; thereforeBC:CFin the duplicate ratio ofBC:GH[V. Def.x.]; and since the figuresBCD,GHIare similar,BCD:GHIin the duplicate ratio ofBC:GH[xx.]; alsoBC:CF:: rectangleBE: rectangleEF. Hence rectangleBE:EF:: figureBCD:GHI; but the rectangleBEis equal to the figureBCD; therefore the rectangleEFis equal to the figureGHI; butEFis equal toA(const.).Thereforethe figureGHIis equal toA, and it is similar toBCD. Hence it is the figurerequired.
Or thus:Describe the squaresEFJK,LMNOequal to the figuresBCDandArespectively[II.xiv.]; then findGHa fourth proportional toEF,LM, andBC[xii.]. OnGHdescribe therectilineal figureGHIsimilar to the figureBCD[xviii.], so thatBCandGHmay be homologoussides.GHIis the figure required.
Dem.—BecauseEF:LM::BC:GH(const.), the figureEFJK:LMNO::BCD:GHI[xxii.]; butEFJKis equal toBCD(const.); thereforeLMNOis equal toGHI; butLMNOisequal toA(const.).ThereforeGHIis equal toA, and it is similar toBCD.
PROP.XXVI.—Theorem.If two similar and similarly situated parallelograms(AEFG,ABCD)have acommon angle, they are about the same diagonal.
PROP.XXVI.—Theorem.If two similar and similarly situated parallelograms(AEFG,ABCD)have acommon angle, they are about the same diagonal.
Dem.—Draw the diagonals (seefig., Prop.xxiv.)AF,AC. Then because the parallelogramsAEFG,ABCDare similar figures, they can be divided into the same number of similar triangles [xx.]. Hence the triangleFAGis similar toCAD, and therefore the angleFAGis equal to the angleCAD. Hence the lineACmust pass through the pointF, andtherefore the parallelograms are about the samediagonal.
Observation.—Propositionxxvi., being the converse ofxxiv., has evidently been misplaced. Thefollowing would be a simpler enunciation:—“If two homothetic parallelograms have a common angle,they are about the same diagonal.”
PROP.XXVII—Problem.To inscribe in a given triangle(ABC)the maximum parallelogram having acommon angle(B)with the triangle.
PROP.XXVII—Problem.To inscribe in a given triangle(ABC)the maximum parallelogram having acommon angle(B)with the triangle.
Sol.—Bisect the sideACopposite to the angleB, atP: throughPdrawPE,PFparallel to the other sides of the triangle.BPis the parallelogramrequired.
Dem.—Take any other pointDinAC: drawDG,DHparallel to the sides, andCKparallel toAB; produceEP,GDto meetCKinKandJ, and produceHDto meetPKinI.
Now, sinceACis bisected inP,EKis also bisected inP; hence [I.xxxvi.] the parallelogramEOis equal toOK; thereforeEOis greater thanDK; butDKis equal toFD[I.xliii.]; henceEOis greater thanFD. To each addBO, and we have the parallelogramBPgreater thanBD.HenceBPis the maximum parallelogramwhich can be inscribed in the given triangle.
Cor. 1.—The maximum parallelogram exceeds any other parallelogram about the same angle in the triangle, by the area of the similar parallelogram whose diagonal is the line between the middle pointPof the opposite side and the pointD, which is the corner of the other inscribed parallelogram.
Cor. 2.—The parallelograms inscribed in a triangle, and having one angle common with it, are proportional to the rectangles contained by the segments of the sides of the triangle, made by the opposite corners of the parallelograms.
Cor. 3.—The parallelogramAC:GH::AC2:AD.DC.
PROP.XXVIII.—Problem.
PROP.XXVIII.—Problem.
To inscribe in a given triangle(ABC)a parallelogram equal to a given rectilinealfigure(X)not greater than the maximum inscribed parallelogram, and having anangle(B)common with the triangle.
Sol.—Bisect the sideACopposite toB, atP. DrawPF,PEparallel to the sidesAB,BC; then [xxvii.]BPis the maximum parallelogram that can be inscribed in the triangleABC; and ifXbe equal to it, the problem is solved. If not, produceEP, and drawCJparallel toPF; then describe the parallelogramKLMN[xxv.] equal to the difference between the figurePJCFandX, and similar toPJCF, and so that the sidesPJandKLwill be homologous; then cut offPIequal toKL; drawIHparallel toAB, cuttingACinD, and drawDGparallel toBC.BDis the parallelogramrequired.
Dem.—Since the parallelogramsPC,PDare about the same diagonal, they are similar [xxiv.]; butPCis similar toKPT(const.); thereforePDis similar toKN, and (const.) their homologous sides,PIandKL, are equal; hence [xx.]PDis equal toKN. Now,PDis the difference betweenEFandGH[xxvii.Cor. 1], andKNis (const.) the difference betweenPCandX; therefore the difference betweenPCandXis equal to the difference betweenEFandGH; butEFis equal toPC.HenceGHis equal toX.
PROP.XXIX.—Problem.
PROP.XXIX.—Problem.
To escribe to a given triangle(ABC)a parallelogram equal to a given rectilinealfigure(X), and having an angle common with an external angle(B)of thetriangle.
Sol.—The construction is the same as the last, except that, instead of making the parallelogramKNequal to the excess of the parallelogramPCover the rectilineal figureX, we make it equal to their sum; and then makePIequal toKL; drawIHparallel toAB, and the rest of the construction as before.
Dem.—Now it can be proved, as in II.vi., that the parallelogramBDis equal to the gnomonOHJ; that is, equal to the difference between the parallelogramsPDandPC, or the difference (const.) betweenKNandPC; that is (const.), equal toX, andBDis escribed to the triangleABC, and has an angle common with the external angleB.Hence the thing required isdone.
Observation.—The enunciations of the three foregoing Propositions have been altered, in order toexpress them in modern technical language. Some writers recommend the student to omitthem—we think differently. In the form we have given them they are freed from their usualrepulsive appearance. The constructions and demonstrations are Euclid’s, but slightlymodified.
PROP.XXX.—Theorem.To divide a given line(AB)in “extreme and mean ratio.”
PROP.XXX.—Theorem.To divide a given line(AB)in “extreme and mean ratio.”
Sol.—DivideABinC, so that the rectangleAB.BCmay be equal to the square onAC[II.xi.]ThenCis the point required.
Dem.—Because the rectangleAB.BCis equal to the square onAC,AB:AC::AC:BC[xvii.].HenceABis cut in extreme and mean ratio inC[Def.ii.].
Exercises.
Exercises.
1.If the three sides of a right-angled triangle be in continued proportion, the hypotenuse isdivided in extreme and mean ratio by the perpendicular from the right angle on thehypotenuse.
2.In the same case the greater segment of the hypotenuse is equal to the least side of thetriangle.
3.The square on the diameter of the circle described about the triangle formed by the pointsF,H,D(seefig.II.xi.), is equal to six times the square on the lineFD.
PROP.XXXI.—Theorem.
PROP.XXXI.—Theorem.
If any similar rectilineal figure be similarly described on the three sides of aright-angled triangle(ABC), the figure on the hypotenuse is equal to the sum of thosedescribed on the two other sides.
Dem.—Draw the perpendicularCD[I.xii.]. Then becauseABCis a right-angled triangle, andCDis drawn from the right angle perpendicular to the hypotenuse;BD:ADin the duplicate ratio ofBA:AC[viii.Cor. 4]. Again, because the figures described onBA,ACare similar, they are in the duplicate ratio ofBA:AC[xx.]. Hence [V.xi.]BA:AD:: figure described onBA: figure described on AC. In like manner,AB:BD:: figure described onAB: figure described onBC. Hence [V.xxiv.]AB: sum ofADandBD:: figure described on the lineAB: sum of the figures described on the linesAC,BC; butABis equal to the sum ofADandBD.Therefore[V.a.]the figure described on the lineABis equal to the sum of the similar figures described on the linesACandBC.
Or thus:Let us denote the sides bya,b,c, and the figures byα,β,γ; then because the figures are similar, we have [xx.]
buta2+b2=c2[I.xlvii.]. Thereforeα+β=γ; that is,the sum of the figures on thesides is equal to the figure on the hypotenuse.
Exercise.
Exercise.
If semicircles be described on supplemental chords of a semicircle, the sum of the areas of thetwo crescents thus formed is equal to the area of the triangle whose sides are the supplementalchords and the diameter.
PROP.XXXII.—Theorem.
PROP.XXXII.—Theorem.
If two triangles(ABC,CDE)which have two sides of one proportional to two sidesof the other(AB:BC::CD:DE), and the contained angles(B,D)equal, bejoined at an angle(C), so as to have their homologous sides parallel, the remainingsides are in the same right line.
Dem.—Because the trianglesABC,CDEhave the anglesBandDequal, and the sides about these angles proportional, viz.,AB:BC::CD:DE, they are equiangular [vi.]; therefore the angleBACis equal toDCE. To each addACD, and we have the sum of the anglesBAC,ACDequal to the sum ofDCEandACD; but the sum ofBAC,ACDis [I.xxix.] two right angles; therefore the sum ofDCEandACDis two right angles.Hence[I.xiv.]AC,CEare in the same rightline.
PROP.XXXIII.–Theorem.
PROP.XXXIII.–Theorem.
In equal circles, angles(BOC,EPF)at the centres or(BAC,EDF)at thecircumferences have the same ratio to one another as the arcs(BC,EF)on whichthey stand, and so also have the sectors(BOC,EPF).
Dem.—1. Take any number of arcsCG,GHin the first circle, each equal toBC. JoinOG,OH, and in the second circle take any number of arcsFI,IJ, each equal toEF. JoinIP,JP. Then because the arcsBC,CG,GHare all equal, the anglesBOC,COG,GOH, are all equal [III.xxvii.]. Therefore the arcBHand the angleBOHare equimultiples of the arcBCand the angleBOC. In like manner it may be proved that the arcEJand the angleEPJare equimultiples of the arcEFand the angleEPF. Again, since the circles are equal, it is evident that the angleBOHis greater than, equal to, or less than the angleEPJ, according as the arcBHis greater than, equal to, or less than the arcEJ. Now we have four magnitudes, namely, the arcBC, the arcEF, the angleBOC, and the angleEPF; and we have taken equimultiples of the first and third, namely, the arcBH, the angleBOH, and other equimultiples of the second and fourth, namely, the arcEJand the angleEPJ, and we have proved that, according as the multiple of the first is greater than, equal to, or less than the multiple of the second, the multiple of the third is greater than, equal to, or less than the multiple of the fourth.Hence[V. Def.v.]BC:EF::the angleBOC:EPF.
Again, since the angleBACis half the angleBOC[III.xx.], andEDFis half the angleEPF,
2. The sectorBOC: sectorEPF::BC:EF.
Dem.—The same construction being made, since the arcBCis equal toCG, the angleBOCis equal toCOG. Hence the sectorsBOC,COGare congruent (seeObservation, Propositionxxix., Book III.); therefore they are equal. In like manner the sectorsCOG,GOHare equal. Hence there are as many equal sectors as there are equal arcs; therefore the arcBHand the sectorBOHare equimultiples of the arcBCand the sectorBOC. In the same manner it may be proved that the arcEJand the sectorEPJare equimultiples of the arcEFand the sectorEPF; and it is evident, by superposition, that if the arcBHis greater than, equal to, or less than the arcEJ, the sectorBOHis greater than, equal to, or less than the sectorEPJ.Hence[V. Def.v.]the arcBC:EF::sectorBOC:sectorEPF.
The second part may be proved as follows:—
SectorBOC=1 2rectangle contained by the arcBC, and the radius of the circleABC[xx.Ex.14] and sectorEPF=1 2rectangle contained by the arcEFand the radius of the circleEDF; and since the circles are equal, their radii are equal. Hence, sectorBOC: sectorEPF:: arcBC: arcEF.