Sol.—JoinCE,CF, and construct a triangleABHonABequiangular toCDE, and similarly placed as regardsCD; that is, make, the angleABHequal toCDE, andBAHequal toDCE. In like manner construct the triangleHAIequiangular toECF, and similarly placed, and lastly, the triangleIAJequiangular and similarly placed withFCG.ThenABHIJis the figurerequired.
Dem.—From the construction it is evident that the figures are equiangular, and it is only required to prove that the sides about the equal angles are proportional. Now because the triangleABHis equiangular toCDE,AB:BH::CD:DE[iv.]; hence the sides about the equal anglesBandDare proportional. Again, from the same triangles we haveBH:HA::DE:EC, and from the trianglesIHA,FEC;HA:HI::EC:EF; therefore (ex æquali)BH:HI::DE:EF; that is, the sides about the equal anglesBHI,DEFare proportional, and so in like manner are the sides about the other equal angles.Hence(Def.i.)the figures aresimilar.
Observation.—In the foregoing construction, the lineABis homologous toCD, and it is evidentthat we may takeABto be homologous to any other side of the given figureCDEFG. Again, ineach case, it the figureABHIJbe turned round the lineABuntil it falls on the other side, it willstill be similar to the figureCDEFG.Hence on a given lineABthere can be constructed two figureseach similar to a given figureCDEFG, and having the given lineABhomologous to any given sideCDof the given figure.
The first of the figures thus constructed is said to bedirectlysimilar, and the secondinverselysimilarto the given figure. These technical terms are due to Hamilton:see“Elements ofQuaternions,” page112.
Cor.1.—Twice as many polygons may be constructed onABsimilar to a given polygonCDEFGas that figure has sides.
Cor.2.—If the figureABHIJbe applied toCDEFGso that the pointAwill coincide withC, and that the lineABmay be placed alongCD, then the pointsH,I,Jwill be respectively on the linesCE,CF,CG; also the sidesBH,HI,IJof the one polygon will be respectively parallel to their homologous sidesDE,EF,FGof the other.
Cor.3.—If lines drawn from any pointOin the plane of a figure to all its angular points be divided in the same ratio, the lines joining the points of division will form a new figure similar to, and having every side parallel to, the homologous side of the original.
PROP.XIX.—Theorem.Similar triangles(ABC,DEF)have their areas to one another in theduplicate ratio of their homologous sides.
PROP.XIX.—Theorem.Similar triangles(ABC,DEF)have their areas to one another in theduplicate ratio of their homologous sides.
Dem.—TakeBGa third proportional toBC,EF[xi.]. JoinAG. Then because the trianglesABC,DEFare similar,AB:BC::DE:EF; hence (alternately)AB:DE::BC:EF; butBC:EF::EF:BG(const.); therefore [V.xi.]AB:DE::EF:BG; hence the sides of the trianglesABG,DEFabout the equal anglesB,Eare reciprocally proportional; therefore the triangles are equal. Again, since the linesBC,EF,BGare continual proportionals,BC:BGin the duplicate ratio ofBC:EF[V. Def.x.]; butBC:BG:: triangleABC:ABG. ThereforeABC:ABGin the duplicate ratio ofBC:EF; but it has been proved that the triangleABGis equal toDEF.Therefore the triangleABCis to the triangleDEFin the duplicate ratio ofBC:EF.
This is the first Proposition in Euclid in which the technical term “duplicate ratio” occurs. My experience with pupils is, that they find it very difficult to understand either Euclid’s proof or his definition. On this account I submit the following alternative proof, which, however, makes use of a new definition of the duplicate ratio of two lines, viz. the ratio of the squares (seeAnnotations on V. Def.x.) described on these lines.
OnABandDEdescribe squares, and throughCandFdraw lines parallel toABandDE, and complete the rectanglesAI,DN.
Now, the trianglesJAC,ODFare evidently equiangular.
Exercises.
Exercises.
1.If one of two similar triangles has its sides 50 per cent.longer than the homologous sides ofthe other; what is the ratio of their areas?
2.When the inscribed and circumscribed regular polygons of any common number of sides to acircle have more than four sides, the difference of their areas is less than the square of the side of theinscribed polygon.
PROP.XX.—Theorem.
PROP.XX.—Theorem.
Similar polygons may be divided(1)into the same number of similar triangles;(2)the corresponding triangles have the same ratio to one another which the polygonshave;(3)the polygons are to each other in the duplicate ratio of their homologoussides.
Dem.—LetABHIJ,CDEFGbe the polygons, and let the sidesAB,CDbe homologous. JoinAH,AI,CE,CF.
1. The triangles into which the polygons are divided are similar. For, since the polygons are similar, they are equiangular, and have the sides about their equal angles proportional [Def.i.]; hence the angleBis equal toD, andAB:BH::CD:DE; therefore [vi.] the triangleABHis equiangular toCDE; hence the angleBHAis equal toDEC; butBHIis equal toDEF(hyp.); therefore the angleAHIis equal toCEF. Again, because the polygons are similar,IH:HB::FE:ED; and since the trianglesABH,CDEare similar,HB:HA::ED:EC; hence (ex aequali)IH:HA::FE:EC, and the angleIHAhas been proved to be equal to the angleFEC;therefore the trianglesIHA,FECare equiangular. In the same manner it can be proved thatthe remaining trianglesare equiangular.
2. Since the triangleABHis similar toCDE, we have [xix.].
ABH : CDE in the duplicate ratio of AH : CE.
In like manner,
In these equal ratios, the trianglesABH,AHI,AIJare the antecedents, and the trianglesCDE,CEF,CFGthe consequents, and [V.xii.] any one of these equal ratios is equal to the ratio of the sum of all the antecedents to the sum of all the consequents;therefore the triangleABH:the triangleCDE::the polygonABHIJ:the polygonCDEFG.
3. The triangleABH:CDEin the duplicate ratio ofAB:CD[xix.].Hence(2)the polygonABHIJ:the polygonCDEFGin the duplicate ratio ofAB:CD.
Cor.1.—The perimeters of similar polygons are to one another in the ratio of their homologous sides.
Cor.2.—As squares are similar polygons, therefore the duplicate ratio of two lines is equal to the ratio of the squares described on them (compare Annotations, V. Def.x.).
Cor.3.—Similar portions of similar figures bear the same ratio to each other as the wholes of the figures.
Cor.4.—Similar portions of the perimeters of similar figures are to each other in the ratio of the whole perimeters.
Exercises.
Exercises.
Def.i.—Homologous pointsin the planes of two similar figures are such, that lines drawn from them to the angular points of the two figures are proportional to the homologous sides of the two figures.
1.If two figures be similar, to each point in the plane of one there will be a corresponding pointin the plane of the other.
Dem.—LetABCD,A′B′C′D′be the two figures,Pa point in the plane ofABCD. JoinAP,BP, and construct a triangleA′P′B′onA′B′, similar toAPB; then it is easy to see that lines fromP′to the angular points ofA′B′C′D′are proportional to the lines fromPto the angular points ofABCD.
2.If two figures be directly similar, and in the same plane, there is in the plane a special pointwhich, regarded as belonging to either figure, is its own homologous point with respect to the other.For, letAB,A′B′be two homologous sides of the figures,Ctheir point of intersection. Through thetwo triads of pointsA,A′,C;B,B′,Cdescribe two circles intersecting again in the pointO:Owillbe the point required. For it is evident that the trianglesOAB,OA′B′are similar andthat either may be turned round the pointO, so that the two bases,AB,A′B′, will beparallel.
Def.ii.—The pointOis called thecentre of similitudeof the figures. It is alsocalled their double point.
3.Two regular polygons ofnsides each havencentres of similitude.
4.If any number of similar triangles have their corresponding vertices lying on three given lines,they have a common centre of similitude.
5.If two figures be directly similar, and have a pair of homologous sides parallel, every pair ofhomologous sides will be parallel.
Def.iii.—Two figures, such as those in 5, are said to behomothetic.
6.If two figures be homothetic, the lines joining corresponding angular points are concurrent,and the point of concurrence is the centre of similitude of the figures.
7.If two polygons be directly similar, either may be turned round their centre of similitude untilthey become homothetic, and this may be done in two different ways.
8.Two circles are similar figures.
Dem.—LetO,O′be their centres; let the angleAOBbe indefinitely small, so that the arcABmay be regarded as a right line; make the angleA′O′B′equal toAOB; then the trianglesAOB,A′O′B′are similar.
Again, make the angleBOCindefinitely small, and makeB′O′C′equal to it; thetrianglesBOC,B′O′C′are similar. Proceeding in this way, we see that the circles can bedivided into the same number of similar elementary triangles. Hence the circles are similarfigures.
9.Sectors of circles having equal central angles are similar figures.
10.As any two points of two circles may be regarded as homologous, two circles have inconsequence an infinite number of centres of similitude; their locus is the circle, whose diameter isthe line joining the two points for which the two circles are homothetic.
11.The areasof circles are to one another as the squares of their diameters. For they are to oneanother as the similar elementary triangles into which they are divided, and these are as the squaresof the radii.
12.The circumferences of circles are as their diameters (Cor.1).
13.The circumference of sectors having equal central angles are proportional to their radii.Hence ifa,a′denote the arcs of two sectors, which subtend equal angles at the centres, and ifr,r′be their radii,a r=a′ r′.
14.The area of a circle is equal to half the rectangle contained by the circumferenceand the radius. This is evident by dividing the circle into elementary triangles, as inEx.8.
15.The area of a sector of a circle is equal to half the rectangle contained by the arc of thesector and the radius of the circle.
PROP.XXI.—Theorem.Rectilineal figures(A,B), which are similar to the same figure(C), aresimilar to one another.
PROP.XXI.—Theorem.Rectilineal figures(A,B), which are similar to the same figure(C), aresimilar to one another.
Dem.—Since the figuresAandCare similar, they are equiangular, and have the sides about their equal angles proportional. In like mannerBandCare equiangular, and have the sides about their equal angles proportional. HenceAandBare equiangular, and have the sides about their equal angles proportional.Therefore theyare similar.
Cor.—Two similar rectilineal figures which are homothetic to a third are homothetic to one another.
Exercise.
Exercise.
If three similar rectilineal figures be homothetic, two by two, their three centres of similitudesare collinear.
PROP.XXII—Theorem.
PROP.XXII—Theorem.
If four lines(AB,CD,EF,GH)be proportional, and any pair of similarrectilineal figures(ABK,CDL)be similarly described on the first and second, andalso any pair(EI,GJ)on the third and fourth, these figures are proportional.Conversely, if any rectilineal figure described on the first of four right lines: thesimilar and similarly described figure described on the second::any rectilineal figureon the third:the similar and similarly described figure on the fourth, the four linesare proportional.
Dem.1.—ABK:CDL::AB2:CD2. [xx.];
IfABK:CDL::EI:GJ,AB:CD::EF:GH.
The enunciation of this Proposition is wrongly stated in Simson’sEuclid, and in those that copy it. As given in those works, the four figures should be similar.
PROP.XXIII.—Theorem.Equiangular parallelograms(AD,CG)are to each other as the rectanglescontained by their sides about a pair of equal angles.
PROP.XXIII.—Theorem.Equiangular parallelograms(AD,CG)are to each other as the rectanglescontained by their sides about a pair of equal angles.
Dem.—Let the two sidesAB,BCabout the equal anglesABD,CBG, be placed so as to form one right line; then it is evident, as in Prop.xiv., thatGB,BDform one right line. Complete the parallelogramBF. Now, denoting the parallelogramsAB,BF,CGbyX,Y,Z, respectively, we have—
Observation.—SinceAB.BD:BC.BGis compounded of the two ratiosAB:BCandBD:BG[V.Def.of compound ratio], the enunciation is the same as if we said, “in the ratio compounded ofthe ratios of the sides,” which is Euclid’s; but it is more easily understood as we have putit.