Chapter 5

Dem.—IfACbe not greater thanAB, it must be either equal to it or less than it. Let us examine each case:—

1. IfACwere equal toAB, the triangleACBwould be isosceles, and then the angleBwould be equal toC[v.]; but it is not by hypothesis; thereforeABis not equal toAC.

2. IfACwere less thanAB, the angleBwould be less than the angleC[xviii.]; but it is not by hypothesis; thereforeACis not less thanAB; and sinceACis neither equal toABnor less than it,it must be greater.

Exercises.

1.Prove this Proposition by a direct demonstration.

2.A line from the vertex of an isosceles triangle to any point in the base is less than either ofthe equal sides, but greater if the point be in the base produced.

3.Three equal lines could not be drawn from the same point to the same line.

4.The perpendicular is the least line which can be drawn from a given point to a given line; andof all others that may be drawn to it, that which is nearest to the perpendicular is less than any onemore remote.

5.If in the fig., Prop.xvi.,ABbe the greatest side of the△ABC,BFis the greatest side ofthe△FBC, and the angleBFCis less than half the angleABC.

6.IfABCbe a△havingABnot greater thanAC, a lineAG, drawn fromAto any pointGinBC, is less thanAC. For the angleACB[xviii.] is not greater thanABC; butAGC[xvi.]is greater thanABC; thereforeAGCis greater thanACG.HenceACis greater thanAG.

PROP. XX.—Theorem.The sum of any two sides(BA,AC)of a triangle(ABC)is greater than thethird.

Dem.—ProduceBAtoD(Post.ii.), and makeADequal toAC[iii.]. JoinCD. Then becauseADis equal toAC, the angleACDis equal toADC(v.); therefore the angleBCDis greater than the angleBDC; hence the sideBDopposite to the greater angle is greater thanBCopposite to the less [xix.]. Again, sinceACis equal toAD, addingBAto both, we have the sum of the sidesBA,ACequal toBD.Therefore the sum ofBA,ACis greater thanBC.

Or thus: Bisect the angleBACbyAE[ix.] Then the angleBEAis greater thanEAC; butEAC=EAB(const.); therefore the angleBEAis greater thanEAB. HenceABis greater thanBE[xix.]. In like mannerACis greater thanEC.Therefore the sum ofBA,ACis greater thanBC.

Exercises.

1.In any triangle, the difference between any two sides is less than the third.

2.If any point within a triangle be joined to its angular points, the sum of the joining lines isgreater than its semiperimeter.

3.If through the extremities of the base of a triangle, whose sides are unequal, lines be drawn toany point in the bisector of the vertical angle, their difference is less than the difference of thesides.

4.If the lines be drawn to any point in the bisector of the external vertical angle, their sum isgreater than the sum of the sides.

5.Any side of any polygon is less than the sum of the remaining sides.

6.The perimeter of any triangle is greater than that of any inscribed triangle, and less thanthat of any circumscribed triangle.

7.The perimeter of any polygon is greater than that of any inscribed, and less than that of anycircumscribed, polygon of the same number of sides.

8.The perimeter of a quadrilateral is greater than the sum of its diagonals.

Def.—A line drawn from any angle of a triangle to the middle point of the opposite side iscalled a medianof the triangle.

9.The sum of the three medians of a triangle is less than its perimeter.

10.The sum of the diagonals of a quadrilateral is less than the sum of the lineswhich can be drawn to its angular points from any point except the intersection of thediagonals.

PROP. XXI.—Theorem.

If two lines(BD,CD)be drawn to a point(D)within a triangle from theextremities of its base(BC), their sum is less than the sum of the remaining sides(BA,CA), but they contain a greater angle.

Dem.—1. ProduceBD(Post.ii.) to meetACinE. Then, in the triangleBAE, the sum of the sidesBA,AEis greater than the sideBE[xx.]: to each addEC, and we have the sum ofBA,ACgreater than the sum ofBE,EC. Again, the sum of the sidesDE,ECof the triangleDECis greater thanDC: to each addBD, and we get the sum ofBE,ECgreater than the sum ofBD,DC; but it has been proved that the sum ofBA,ACis greater than the sum ofBE,EC.Therefore much more is the sum ofBA,ACgreater than the sum ofBD,DC.

2. The external angleBDCof the triangleDECis greater than the internal angleBEC[xvi.], and the angleBEC, for a like reason, is greater thanBAC.Therefore much more isBDCgreater thanBAC.

Part 2 may be proved without producing either of the sidesBD,DC. Thus: joinADand produce it to meetBCinF; then the angleBDFis greater than the angleBAF[xvi.], andFDCis greater thanFAC.Therefore the whole angleBDCisgreater thanBAC.

Exercises.

1.The sum of the lines drawn from any point within a triangle to its angular points is less thanthe perimeter. (Compare Ex.2, last Prop.)

2.If a convex polygonal lineABCDlie within a convex polygonal lineAMNDterminating inthe same extremities, the length of the former is less than that of the latter.

PROP.XXII.—Problem.

To construct a triangle whose three sides shall be respectively equal to threegiven lines(A,B,C), the sum of every two of which is greater than thethird.

Sol.—Take any right lineDE, terminated atD, but unlimited towardsE, and cut off [iii.]DFequal toA,FGequal toB, andGHequal toC. WithFas centre, andFDas radius, describe the circleKDL(Post.iii.); and withGas centre, andGHas radius, describe the circleKHL, intersecting the former circle inK. JoinKF,KG.KFGis the triangle required.

Dem.—SinceFis the centre of the circleKDL,FKis equal toFD; butFDis equal toA(const.); therefore (Axiomi.)FKis equal toA. In like mannerGKis equal toC, andFGis equal toB(const.)Hence the threesides of the triangleKFGare respectively equal to the three linesA,B,C.

Questions for Examination.

1.What is the reason for stating in the enunciation that the sum of every two of the given linesmust be greater than the third?

2.Prove that when that condition is fulfilled the two circles must intersect.

3.Under what conditions would the circles not intersect?

4.If the sum of two of the lines were equal to the third, would the circles meet? Prove that theywould not intersect.

PROP.XXIII.—Problem.At a given point(A)in a given right line(AB)to make an angle equal to agiven rectilineal angle(DEF).

Sol.—In the sidesED,EFof the given angle take any arbitrary pointsDandF. JoinDF, and construct [xxii.] the triangleBAC, whose sides, taken in order, shall be equal to those ofDEF—namely,ABequal toED,ACequal toEF, andCBequal toFD; then the angleBACwill [viii.] be equal toDEF.Hence it is therequired angle.

Exercises.

1.Construct a triangle, being given two sides and the angle between them.

2.Construct a triangle, being given two angles and the side between them.

3.Construct a triangle, being given two sides and the angle opposite to one of them.

4.Construct a triangle, being given the base, one of the angles at the base, and the sum ordifference of the sides.

5.Given two points, one of which is in a given line, it is required to find another point in thegiven line, such that the sum or difference of its distances from the former points may be given.Show that two such points may be found in each case.

PROP.XXIV.—Theorem.

If two triangles(ABC,DEF)have two sides(AB,AC)of one respectively equalto two sides(DE,DF)of the other, but the contained angle(BAC)of one greaterthan the contained angle(EDF)of the other, the base of that which has the greaterangle is greater than the base of the other.

Dem.—Of the two sidesAB,AC, letABbe the one which is not the greater, and with it make the angleBAGequal toEDF[xxiii.]. Then becauseABis not greater thanAC,AGis less thanAC[xix., Exer. 6]. ProduceAGtoH, and makeAHequal toDForAC[iii.]. JoinBH,CH.

In the trianglesBAH,EDF, we haveABequal toDE(hyp.),AHequal toDF(const.), and the angleBAHequal to the angleEDF(const.); therefore the base [iv.]BHis equal toEF. Again, becauseAHis equal toAC(const.), the triangleACHis isosceles; therefore the angleACHis equal toAHC[v.]; butACHis greater thanBCH; thereforeAHCis greater thanBCH: much more is the angleBHCgreater thanBCH, and the greater angle is subtended by the greater side [xix.]. ThereforeBCis greater thanBH; butBHhas been proved to be equal toEF;thereforeBCis greater thanEF.

The concluding part of this Proposition may be proved without joiningCH, thus:—

BG+GH > BH[xx.],AG+GC > AC[xx.];thereforeBC+AH > BH+AC;butAH=AC(const.);thereforeBCis> BH.

BCis> BH.

Or thus:Bisect the angleCAHbyAO. JoinOH. Now in the△sCAO,HAOwe have the sidesCA,AOin one equal to the sidesAH,AOin the other, and the contained angles equal; thereforethe baseOCis equal to the baseOH[iv.]: to each addBO, and we haveBCequal to the sum ofBO,OH; but the sum ofBO,OHis greater thanBH[xx.].ThereforeBCis greater thanBH, thatis, greater thanEF.

Exercises.

1.Prove this Proposition by making the angleABHto the left ofAB.

2.Prove that the angleBCAis greater thanEFD.

PROP.XXV.—Theorem.

If two triangles(ABC,DEF)have two sides(AB,AC)of one respectively equalto two sides(DE,DF)of the other, but the base(BC)of one greater than the base(EF)of the other, the angle(A)contained by the sides of that which hasthe greater base is greater them the angle(D)contained by the sides of theother.

Dem.—If the angleAbe not greater thanD, it must be either equal to it or less than it. We shall examine each case:—

1. IfAwere equal toD, the trianglesABC,DEFwould have the two sidesAB,ACof one respectively equal to the two sidesDE,DFof the other, and the angleAcontained by the two sides of one equal to the angleDcontained by the two sides of the other. Hence [iv.]BCwould be equal toEF; butBCis, by hypothesis, greater thanEF; hence the angleAis not equal to the angleD.

2. IfAwere less thanD, thenDwould be greater thanA, and the trianglesDEF,ABCwould have the two sidesDE,DFof one respectively equal to the two sidesAB,ACof the other, and the angleDcontained by the two sides of one greater than the angleAcontained by the two sides of the other. Hence [xxiv.]EFwould be greater thanBC; butEF(hyp.) is not greater thanBC. ThereforeAis not less thanD, and we have proved that it is not equal to it;therefore it must begreater.

Or thus, directly: Construct the triangleACG, whose three sidesAG,GC,CAshall be respectively equal to the three sidesDE,EF,FDof the triangleDEF[xxii.]. JoinBG. Then becauseBCis greater thanEF,BCis greater thanCG. Hence [xviii.] the angleBGCis greater thanGBC; and make (xxiii.) the angleBGHequal toGBH, and joinAH. Then [vi.]BHis equal toGH. Therefore the trianglesABH,AGHhave the sidesAB,AHof one equal to the sidesAG,AHof the other, and the baseBHequal toGH. Therefore [viii.] the angleBAHis equal toGAH.Hence the angleBACis greater thanCAG, and therefore greater thanEDF.

Exercise.

Demonstrate this Proposition directly by cutting off fromBCa part equal toEF.

PROP.XXVI.—Theorem.

If two triangles(ABC,DEF)have two angles(B,C)of one equal respectively totwo angles(E,F)of the other, and a side of one equal to a side similarly placedwith respect to the equal angles of the other, the triangles are equal in everyrespect.

Dem.—This Proposition breaks up into two according as the sides given to be equal are the sides adjacent to the equal angles, namelyBCandEF, or those opposite equal angles.

1. Let the equal sides beBCandEF; then ifDEbe not equal toAB, supposeGEto be equal to it. JoinGF; then the trianglesABC,GEFhave the sidesAB,BCof one respectively equal to the sidesGE,EFof the other, and the angleABCequal to the angleGEF(hyp.); therefore [iv.] the angleACBis equal to the angleGFE; but the angleACBis (hyp.) equal toDFE; henceGFEis equal toDFE—a part equal to the whole, which is absurd; thereforeABandDEare not unequal, that is, they are equal. Consequently the trianglesABC,DEFhave the sidesAB,BCof one respectively equal to the sidesDE,EFof the other; and the contained anglesABCandDEFequal;therefore[iv.]ACis equal toDF, and the angleBACis equal to the angleEDF.

2. Let the sides given to be equal beABandDE; it is required to prove thatBCis equal toEF, andACtoDF. IfBCbe not equal toEF, supposeBGto be equal to it. JoinAG. Then the trianglesABG,DEFhave the two sidesAB,BGof one respectively equal to the two sidesDE,EFof the other, and the angleABGequal to the angleDEF; therefore [iv.] the angleAGBis equal toDFE; but the angleACBis equal toDFE(hyp.). Hence (Axiomi.) the angleAGBis equal toACB, that is, the exterior angle of the triangleACGis equal to the interior and non-adjacent angle, which [xvi.] is impossible.HenceBCmust be equal toEF, and thesame as in1,ACis equal toDF, and the angleBACis equal to the angleEDF.

This Proposition, together withiv.andviii., includes all the cases of the congruence of twotriangles. PartI. may be proved immediately by superposition. For it is evident ifABCbe appliedtoDEF, so that the pointBshall coincide withE, and the lineBCwithEF, sinceBCis equal toEF, the pointCshall coincide withF; and since the anglesB,Care respectively equal to theanglesE,F, the linesBA,CAshall coincide withEDandFD.Hence the triangles arecongruent.

Def.—If every point on a geometrical figure satisfies an assigned condition, thatfigure is called the locusof the point satisfying the condition.Thus, for example, a circle is the locus of a point whose distance from the centre is equal to its radius.

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