PROP.I.—Problem.On a given finite right line(AB)to construct an equilateral triangle.
PROP.I.—Problem.On a given finite right line(AB)to construct an equilateral triangle.
Sol.—WithAas centre, andABas radius, describe the circleBCD(Post.iii.). WithBas centre, andBAas radius, describe the circleACE, cutting the former circle inC. JoinCA,CB(Post.i.). ThenABCis the equilateral trianglerequired.
Dem.—BecauseAis the centre of the circleBCD,ACis equal toAB(Def.xxxii.). Again, becauseBis the centre of the circleACE,BCis equal toBA. Hence we have proved.
But things which are equal to the same are equal to one another (Axiomi.); thereforeACis equal toBC; therefore the three linesAB,BC,CAare equal to one another. Hence the triangleABCis equilateral (Def.xxi.); and it is described on the given lineAB,which was required to be done.
Questions for Examination.
Questions for Examination.
1.What is thedatumin this proposition?
2.What is thequaesitum?
3.What is a finite right line?
4.What is the opposite of finite?
5.In what part of the construction is the third postulate quoted? and for what purpose? Whereis the first postulate quoted?
6.Where is the first axiom quoted?
7.What use is made of the definition of a circle? What is a circle?
8.What is an equilateral triangle?
Exercises.
Exercises.
The following exercises are to be solved when the pupil has mastered the First Book:—
1.If the linesAF,BFbe joined, the figureACBFis a lozenge.
2.IfABbe produced toDandE, the trianglesCDFandCEFare equilateral.
3.IfCA,CBbe produced to meet the circles again inGandH, the pointsG,F,Harecollinear, and the triangleGCHis equilateral.
4.IfCFbe joined,CF2= 3AB2.
5.Describe a circle in the spaceACB, bounded by the lineABand the two circles.
PROP.II.—Problem.From a given point(A)to draw a right line equal to a given finite right line(BC).
PROP.II.—Problem.From a given point(A)to draw a right line equal to a given finite right line(BC).
Sol.—JoinAB(Post.i.); onABdescribe the equilateral triangleABD[i.]. WithBas centre, andBCas radius, describe the circleECH(Postiii.). ProduceDBto meet the circleECHinE(Post.ii.). WithDas centre, andDEas radius, describe the circleEFG(Post.iii.). ProduceDAto meet this circle inF.AFis equal toBC.
Dem.—BecauseDis the centre of the circleEFG,DFis equal toDE(Def.xxxii.). And becauseDABis an equilateral triangle,DAis equal toDB(Def.xxi.). Hence we have
and taking the latter from the former, the remainderAFis equal to the remainderBE(Axiomiii.). Again, becauseBis the centre of the circleECH,BCis equal toBE; and we have proved thatAFis equal toBE; and things which are equal to the same thing are equal to one another (Axiomi.). HenceAFis equal toBC.Therefore from the given pointAthe lineAFhas been drawn equal toBC.
It is usual with commentators on Euclid to say that he allows the use of theruleandcompass.Were such the case this Proposition would have been unnecessary. The fact is, Euclid’s object was toteach Theoretical and not Practical Geometry, and the only things he postulates are thedrawing of right lines and the describing of circles. If he allowed the mechanical use of therule and compass he could give methods of solving many problems that go beyond thelimits of the “geometry of the point, line, and circle.”—SeeNotes D, F at the end of thiswork.
Exercises.
Exercises.
1.Solve the problem when the pointAis in the lineBCitself.
2.Inflect from a given pointAto a given lineBCa line equal to a given line. State the numberof solutions.
PROP. III.—Problem.From the greater(AB)of two given right lines to cut off a part equal to(C)the less.
PROP. III.—Problem.From the greater(AB)of two given right lines to cut off a part equal to(C)the less.
Sol.—FromA, one of the extremities ofAB, draw the right lineADequal toC[ii.]; and withAas centre, andADas radius, describe the circleEDF(Post.iii.) cuttingABinE.AEshall be equal toC.
Dem.—BecauseAis the centre of the circleEDF,AEis equal toAD(Def.xxxii.), andCis equal toAD(const.); and things which are equal to the same are equal to one another (Axiomi.); thereforeAEis equal toC.Wherefore fromAB, the greater of the two given lines, a part,AE, has been out off equal toC, theless.
Questions for Examination.
Questions for Examination.
1.What previous problem is employed in the solution of this?
2.What postulate?
3.What axiom in the demonstration?
4.Show how to produce the less of two given lines until the whole produced line becomes equalto the greater.
PROP.IV.—Theorem.
PROP.IV.—Theorem.
If two triangles(BAC,EDF)have two sides(BA,AC)of one equal respectivelyto two sides(ED,DF)of the other, and have also the angles(A,D)includedby those sides equal, the triangles shall be equal in every respect—that is,their bases or third sides(BC,EF)shall be equal, and the angles(B,C)at the base of one shall be respectively equal to the angles(E,F)at thebase of the other; namely, those shall be equal to which the equal sides areopposite.
Dem.—Let us conceive the triangleBACto be applied toEDF, so that the pointAshall coincide withD, and the lineABwithDE, and that the pointCshall be on the same side ofDEasF; then becauseABis equal toDE, the pointBshall coincide withE. Again, because the angleBACis equal to the angleEDF, the lineACshall coincide withDF; and sinceACis equal toDF(hyp.), the pointCshall coincide withF; and we have proved that the pointBcoincides withE. Hence two points of the lineBCcoincide with two points of the lineEF; and since two right lines cannot enclose a space,BCmust coincide withEF. Hence the triangles agree in every respect;thereforeBCis equal toEF, the angleBis equal to theangleE, the angleCto the angleF, and the triangleBACto the triangleEDF.
Questions for Examination.
Questions for Examination.
1.How many parts in the hypothesis of this Proposition?Ans.Three. Name them.
2.How many in the conclusion? Name them.
3.What technical term is applied to figures which agree in everything but position?Ans.Theyare said to be congruent.
4.What is meant by superposition?
5.What axiom is made use of in superposition?
6.How many parts in a triangle?Ans.Six; namely, three sides and three angles.
7.When it is required to prove that two triangles are congruent, how many parts of one mustbe given equal to corresponding parts of the other?Ans.In general, any three exceptthe three angles. This will be established in Props.viii.andxxvi., taken along withiv.
8.What property of two lines having two common points is quoted in this Proposition? Theymust coincide.
Exercises.
Exercises.
1.The line that bisects the vertical angle of an isosceles triangle bisects the baseperpendicularly.
2.If two adjacent sides of a quadrilateral be equal, and the diagonal bisects the angle betweenthem, their other sides are equal.
3.If two lines be at right angles, and if each bisect the other, then any point in either is equallydistant from the extremities of the other.
4.If equilateral triangles be described on the sides of any triangle, the distances between thevertices of the original triangle and the opposite vertices of the equilateral triangles are equal. (ThisProposition should be proved after the student has read Prop.xxxii.)
PROP.V.—Theorem.
PROP.V.—Theorem.
The angles(ABC,ACB)at the base(BC)of an isosceles triangle are equal to oneanother, and if the equal sides(AB,AC)be produced, the external angles(DEC,ECB)below the base shall be equal.
Dem.—InBDtake any pointF, and fromAE, the greater, cut offAGequal toAF[iii]. JoinBG,CF(Post.i.). BecauseAFis equal toAG(const.), andACis equal toAB(hyp.), the two trianglesFAC,GABhave the sidesFA,ACin one respectively equal to the sidesGA,ABin the other; and the included angleAis common to both triangles. Hence [iv.] the baseFCis equal toGB, the angleAFCis equal toAGB, and the angleACFis equal to the angleABG.
Again, becauseAFis equal toAG(const.), andABtoAC(hyp.), the remainder,BF, is equal toCG(Axiomiii); and we have proved thatFCis equal toGB, and the angleBFCequal to the angleCGB. Hence the two trianglesBFC,CGBhave the two sidesBF,FCin one equal to the two sidesCG,GBin the other; and the angleBFCcontained by the two sides of one equal to the angleCGBcontained by the two sides of the other. Therefore [iv.] these triangles have the angleFBCequal to the angleGCB,and these are the angles below thebase. Also the angleFCBequal toGBC; but the whole angleFCAhas been proved equal to the whole angleGBA. Hence the remaining angleACBis equal to the remaining angleABC,and these are the angles at thebase.
Observation.—The great difficulty which beginners find in this Proposition is due to the factthat the two trianglesACF,ABGoverlap each other. The teacher should make thesetriangles separate, as in the annexed diagram, and point out the corresponding partsthus:—
The student should also be shown how to apply one of the triangles to the other, so as tobring them into coincidence. Similar Illustrations may be given of the trianglesBFC,CGB.
The following is a very easy proof of this Proposition. Conceive the△ACBto be turned,without alteration, round the lineAC, until it falls on the other side. LetACDbe its newposition; then the angleADCof the displaced triangle is evidently equal to the angleABC,with which it originally coincided. Again, the two△sBAC,CADhave the sidesBA,ACof one respectively equal to the sidesAC,ADof the other, and the included anglesequal; therefore [iv.] the angleACBopposite to the sideABis equal to the angleADCopposite to the sideAC; but the angleADCis equal toABC; thereforeACBis equal toABC.
Cor.—Every equilateral triangle is equiangular.
Def.—A line in any figure, such asACin the preceding diagram, which is suchthat, by folding the plane of the figure round it, one part of the diagram will coincidewith the other, is called anaxisofsymmetryof the figure.
Exercises.
Exercises.
1.Prove that the angles at the base are equal without producing the sides. Also by producingthe sides through the vertex.
2.Prove that the line joining the pointAto the intersection of the linesCFandBGis an axisof symmetry of the figure.
3.If two isosceles triangles be on the same base, and be either at the same or at oppositesides of it, the line joining their vertices is an axis of symmetry of the figure formed bythem.
4.Show how to prove this Proposition by assuming as an axiom that every angle has abisector.
5.Each diagonal of a lozenge is an axis of symmetry of the lozenge.
6.If three points be taken on the sides of an equilateral triangle, namely, one on eachside, at equal distances from the angles, the lines joining them form a new equilateraltriangle.
PROP.VI.—Theorem.If two angles(B,C)of a triangle be equal, the sides(AC,AB)opposite tothem are also equal.
PROP.VI.—Theorem.If two angles(B,C)of a triangle be equal, the sides(AC,AB)opposite tothem are also equal.
Dem.—IfAB,ACare not equal, one must be greater than the other. SupposeABis the greater, and that the partBDis equal toAC. JoinCD(Post.i.). Then the two trianglesDBC,ACBhaveBDequal toAC, andBCcommon to both. Therefore the two sidesDB,BCin one are equal to the two sidesAC,CBin the other; and the angleDBCin one is equal to the angleACBin the other (hyp). Therefore [iv.] the triangleDBCis equal to the triangleACB—the less to the greater, which is absurd;henceAC,ABare not unequal, that is, they areequal.
Questions for Examination.
Questions for Examination.
1.What is the hypothesis in this Proposition?
2.What Proposition is this the converse of?
3.What is the obverse of this Proposition?
4.What is the obverse of Prop.v.?
5.What is meant by an indirect proof?
6.How does Euclid generally prove converse Propositions?
7.What false assumption is made in the demonstration?
8.What does this assumption lead to?
PROP.VII—Theorem.
PROP.VII—Theorem.
If two triangles(ACB,ADB)on the same base(AB)and on the same side of ithave one pair of conterminous sides(AC,AD)equal to one another, the other pairof conterminous sides(BC,BD)must be unequal.
Dem.—1. Let the vertex of each triangle be without the other. JoinCD. Then becauseADis equal toAC(hyp.), the triangleACDis isosceles; therefore [v.] the angleACDis equal to the angleADC; butADCis greater thanBDC(Axiomix.); thereforeACDis greater thanBDC: much, more isBCDgreater thanBDC. Now if the sideBDwere equal toBC, the angleBCDwould be equal toBDC[v.]; but it has been proved to be greater.HenceBDis not equal toBC.
2. Let the vertex of one triangleADBfall within the other triangleACB. Produce the sidesAC,ADtoEandF. Then becauseACis equal toAD(hyp.), the triangleACDis isosceles, and [v.] the external anglesECD,FDCat the other side of the baseCDare equal; butECDis greater thanBCD(Axiomix.). ThereforeFDCis greater thanBCD: much more isBDCgreater thanBCD; but ifBCwere equal toBD, the angleBDCwould be equal toBCD[v.];thereforeBCcannot beequal toBD.
3. If the vertexDof the second triangle fall on the lineBC, it is evident thatBCandBDare unequal.
Questions for Examination.
Questions for Examination.
1.What use is made of Prop.vii.?Ans.As a lemma to Prop.viii.
2.In the demonstration of Prop.vii.the contrapositive of Prop.v.occurs; showwhere.
3.Show that two circles can intersect each other only in one point on the same side of the linejoining their centres, and hence that two circles cannot have more than two points ofintersection.