PROP.VI.—Theorem.
PROP.VI.—Theorem.
If two triangles(ABC,DEF)have one angle(A)in one equal to one angle(D)inthe other, and the sides about these angles proportional(BA:AC::ED:DF), thetriangles are equiangular, and have those angles equal which are opposite to thehomologous sides.
Dem.—Make the same construction as in the last Proposition; then the trianglesABC,DEGare equiangular.
ThereforeDGis equal toDF. Again, because the angleEDGis equal toBAC(const.), andBACequal toEDF(hyp.), the angleEDGis equal toEDF; and it has been proved thatDGis equal toDF, andDEis common; hence the trianglesEDGandEDFare equiangular; butEDGis equiangular toBAC.ThereforeEDFis equiangular toBAC.
[It is easy to see, as in the case of Propositioniv., that an immediate proof of this Propositioncan also be got from Propositionii.].
Cor.1.—If the ratio of two sides of a triangle be given, and the angle between them, the triangle is given in species.
PROP.VII.—Theorem.
PROP.VII.—Theorem.
If two triangles(ABC,DEF)have one angle(A)one equal to one angle(D)in theother, the sides about two other angles(B,E)proportional(AB:BC::DE:EF),and the remaining angles(C,F)of the same species (i.e. either both acute or bothnot acute), the triangles are similar.
Dem.—If the anglesBandEare not equal, one must be greater than the other. SupposeABCto be the greater, and that the partABGis equal toDEF, then the trianglesABG,DEFhave two angles in one equal to two angles in the other, and are [I.xxxii.] equiangular.
ThereforeBGis equal toBC. Hence the anglesBCG,BGCmust be each acute [I.xvii.]; thereforeAGBmust be obtuse; henceDFE, which is equal to it, is obtuse; and it has been proved thatACBis acute; therefore the anglesACB,DFEare of different species; but (hyp.) they are of the same species, which is absurd. Hence the anglesBandEare not unequal, that is, they are equal.Therefore the triangles areequiangular.
Cor.1.—If two trianglesABC,DEFhave two sides in one proportional to two sides in the other,AB:BC::DE:EF, and the anglesA,Dopposite one pair of homologous sides equal, the anglesC,Fopposite the other are either equal or supplemental. This Proposition is nearly identical withvii.
Cor.2.—If either of the anglesC,Fbe right, the other must be right.
PROP.VIII.—Theorem.
PROP.VIII.—Theorem.
The triangles(ACD,BCD)into which a right-angled triangle(ACB)is divided,by the perpendicular(CD)from the right angle(C)on the hypotenuse, are similar tothe whole and to one another.
Dem.—Since the two trianglesADC,ACBhave the angleAcommon, and the anglesADC,ACBequal, each being right, they are [I.xxxii.] equiangular; hence [iv.] they are similar. In like manner it may be proved thatBDCis similar toABC. HenceADC,CDBare each similar toACD, and therefore they are similar to oneanother.
Cor.1.—The perpendicularCDis a mean proportional between the segmentsAD,DBof the hypotenuse.
For, since the trianglesADC,CDBare equiangular, we haveAD:DC::DC:DB; henceDCis a mean proportional betweenAD,DB(Def.iii.).
Cor.2.—BCis a mean proportional betweenAB,BD; andACbetweenAB,AD.
Cor.3.—The segmentsAD,DBare in the duplicate ofAC:CB, or in other words,AD:DB::AC2:CB2,
Cor.4.—BA:ADin the duplicate ratios ofBA:AC; andAB:BDin the duplicate ratio ofAB:BC.
PROP.IX.—Problem.From a given right line(AB)to cut off any part required (i.e. to cut off anyrequired submultiple)
PROP.IX.—Problem.From a given right line(AB)to cut off any part required (i.e. to cut off anyrequired submultiple)
.
Sol.—Let it be required, for instance, to cut off the fourth part. DrawAF, making any angle withAB, and inAFtake any pointC, and cut off (I.iii.) the partsCD,DE,EFeach equal toAC. JoinBF, and drawCGparallel toBF.AGis the fourth part ofAB.
Dem.—SinceCGis parallel to the sideBFof the triangleABF,AC:AF::AG:AB[ii.]; butACis the fourth part ofAF(const.). HenceAGis the fourth part ofAB[V.,d.].In the same manner, any other required submultiple maybe cut off.
Propositionx., Book I., is a particular case of this Proposition.
PROP.X.—Problem.To divide a given undivided line(AB)similarly to a given divided line(CD).
PROP.X.—Problem.To divide a given undivided line(AB)similarly to a given divided line(CD).
Sol.—DrawAG, making any angle withAB, and cut off the partsAH,HI,IGrespectively equal to the partsCE,EF,FDof the given divided lineCD. JoinBG, and drawHK,IL, each parallel toBG.ABwill be divided similarly toCD.
Dem.—ThroughHdrawHNparallel toAB, cuttingILinM. Now in the triangleALI,HKis parallel toIL. Hence [ii.]AK:KL::AH:HI, that is ::CE:EF(const.). Again, in the triangleHNG,MIis parallel toNG. Therefore [ii.]HM:MN::HI:IG; but [I.xxxiv.]HMis equal toKL,MNis equal toLB,HIis equal toEF, andIGis equal toFD(const.). ThereforeKL:LB::EF:FD.Hence the lineABis divided similarly to the lineCD.
Exercises.
Exercises.
1.To divide a given lineABinternallyorexternallyin the ratio of two given lines,m,n.
Sol.—ThroughAandBdraw any two parallelsACandBDin opposite directions. Cut offAC=m, andBD=n, and joinCD; the joining line will divideABinternally atEin the ratio ofm:n.
2.IfBD′be drawn in the same direction withAC, as denoted by the dotted line, thenCD′willcutABexternally atE′in the ratio ofm:n.
Cor.—The two pointsE,E′divideABharmonically.
This problem is manifestly equivalent to the following:—Given the sum or difference of two linesand their ratio, to find the lines.
3.Any lineAE′, through the middle pointBof the baseDD′of a triangleDCD′,is cut harmonically by the sides of the triangle and a parallel to the base through thevertex.
4.Given the sum of the squares on two lines and their ratio; find the lines.
5.Given the difference of the squares on two lines and their ratio; find the lines.
6.Given the base and ratio of the sides of a triangle; construct it when any of the following datais given:—1, the area; 2, the difference on the squares of the sides; 3, the sum of the squares on thesides; 4, the vertical angle; 5, the difference of the base angles.
PROP.XI.—Problem.To find a third proportionalto two given lines(X,Y).
PROP.XI.—Problem.To find a third proportionalto two given lines(X,Y).
Sol.—Draw any two linesAC,AEmaking an angle. Cut offABequalX,BCequalY, andADequalY. JoinBD, and drawCEparallel toBD, thenDEis the third proportional required.
Dem.—In the triangleCAE,BDis parallel toCE; thereforeAB:BC::AD:DE[ii.]; butABis equal toX, andBC,ADeach equal toY. ThereforeX:Y::Y:DE.HenceDEis a third proportional toXandY.
Another solution can be inferred from Propositionviii.For ifAD,DCin that Proposition berespectively equal toXandY, thenDBwill be the third proportional. Or again, if in the diagram,Propositionviii.,AD=X, andAC=Y,ABwill be the third proportional. Hence may be inferreda method of continuing the proportion to any number of terms.
Exercises.
Exercises.
1.IfAOΩ be a triangle, having the sideAΩ greater thanAO; then if we cut offAB=AO, drawBB′parallel toAO, cut offBC=BB′, &c., the series of linesAB,BC,CD, &c., are in continualproportion.
2.AB−BC:AB::AB:AΩ. This is evident by drawing throughB′a parallel toAΩ.
PROP.XII.—Problem.To find a fourth proportionalto three given lines(X,Y,Z).
PROP.XII.—Problem.To find a fourth proportionalto three given lines(X,Y,Z).
Sol.—Draw any two linesAC,AE, making an angle; then cut offABequalX,BCequalY,ADequalZ. JoinBD, and drawCEparallel toBD.DEwill be thefourth proportional required.
Dem.—SinceBDis parallel toCE, we have [ii.]AB:BC::AD:DE; thereforeX:Y::Z:DE.HenceDEis a fourth proportional toX,Y,Z.
Or thus:Take two linesAD,BCintersecting inO. MakeOA=X,OB=Y,OC=Z, and describe a circle through the pointsA,B,C[IV.v.] cuttingADinD.ODwill be the fourth proportional required.
The demonstration is evident from the similarity of the trianglesAOBandCOD.
PROP.XIII.—Problem.To find a mean proportionalbetween two given lines.(X,Y).
PROP.XIII.—Problem.To find a mean proportionalbetween two given lines.(X,Y).
Sol.—Take on any lineACpartsAB,BCrespectively equal toX,Y. OnACdescribe a semicircleADC. ErectBDat right angles toAC, meeting the semicircle inD.BDwill be the mean proportional required.
Dem.—JoinAD,DC. SinceADCis a semicircle, the angleADCis right [III.xxxi.]. Hence, sinceADCis a right-angled triangle, andBDa perpendicular from the right angle on the hypotenuse,BDis a mean proportional [viii.Cor.1] betweenAB,BC; that is,BDis a mean proportional betweenXandY.
Exercises.
Exercises.
1.Another solution may be inferred from Propositionviii.,Cor.2.
2.If through any point within a circle the chord be drawn, which is bisected in that point, itshalf is a mean proportional between the segments of any other chord passing through the samepoint.
3.The tangent to a circle from any external point is a mean proportional between the segmentsof any secant passing through the same point.
4.If through the middle pointCof any arc of a circle any secant be drawn cutting the chord ofthe arc inD, and the circle again inE, the chord of half the arc is a mean proportional betweenCDandCE.
5.If a circle be described touching another circle internally and two parallel chords, theperpendicular from the centre of the former on the diameter of the latter, which bisects the chords,is a mean proportional between the two extremes of the three segments into which the diameter isdivided by the chords.
6.If a circle be described touching a semicircle and its diameter, the diameter of the circle is aharmonic mean between the segments into which the diameter of the semicircle is divided at thepoint of contact.
7.State and prove the Proposition corresponding to Ex.5, for external contact of thecircles.
PROP.XIV.—Theorem.
PROP.XIV.—Theorem.
1. Equiangular parallelograms(AB,CD)which are equal in area have the sidesabout the equal angles reciprocally proportional—AC:CE::GC:CB.2. Equiangular parallelograms which have the sides about the equal anglesreciprocally proportional are equal in area.
Dem.—LetAC,CEbe so placed as to form one right line, and that the equal anglesACB,ECGmay be vertically opposite. Now, since the angleACBis equal toECG, to each addBCE, and we have the sum of the anglesACB,BCEequal to the sum of the anglesECG,BCE; but the sum ofACB,BCEis [I.xiii.] two right angles. Therefore the sum ofECG,BCEis two right angles. Hence [I.xiv.]BC,CGform one right line. Complete the parallelogramBE.
Again, since the parallelogramsAB,CDare equal (hyp.),
that is,the sides about the equal angles are reciprocally proportional.
2. LetAC:CE::GC:CB, to prove the parallelogramsAB,CDare equal.
Dem.—Let the same construction be made, we have
that is,the parallelograms are equal.
Or thus:JoinHE,BE,HD,BD. ThePICTHC= twice the△HBE, and thePICTCD= twice the△BDE. Therefore the△HBE=BDE, and [I.xxxix.]HDis paralleltoBE. Hence
HB :BF ::DE :EF ; that is, AC :CE ::GC :CB.
2.May be proved by reversing this demonstration.
Another demonstration of this Proposition may be got by producing the linesHAandDGto meet inI. Then [I.xliii.] the pointsI,C,Fare collinear, and the Proposition isevident.