Chapter 4

PROP.VIII.—Theorem.

If two triangles(ABC,DEF)have two sides(AB,AC)of one respectivelyequal to two sides(DE,DF)of the other, and have also the base(BC)of one equalto the base(EF)of the other; then the two triangles shall be equal, and the angles ofone shall be respectively equal to the angles of the other—namely, those shall be equalto which the equal sides are opposite.

Dem.—Let the triangleABCbe applied toDEF, so that the pointBwill coincide withE, and the lineBCwith the lineEF; then becauseBCis equal toEF, the pointCshall coincide withF. Then if the vertexAfall on the same side ofEFas the vertexD, the pointAmust coincide withD; for if not, let it take a different positionG; then we haveEGequal toBA, andBAis equal toED(hyp.). Hence (Axiomi.)EGis equal toED: in like manner,FGis equal toFD, and this is impossible [vii.]. Hence the pointAmust coincide withD, and the triangleABCagrees in every respect with the triangleDEF;andtherefore the three angles of one are respectively equal to the three angles of theother—namely,AtoD,BtoE, andCtoF, and the two triangles areequal.

This Proposition is the converse ofiv., and is the second case of the congruence of triangles in the Elements.

Philo’s Proof.—Let the equal bases be applied as in the foregoing proof, but let the vertices beon the opposite sides; then letBGCbe the position whichEDFtakes. JoinAG. Then becauseBG=BA, the angleBAG=BGA. In like manner the angleCAG=CGA. Hence the whole angleBAC=BGC; butBGC=EDFthereforeBAC=EDF.

PROP. IX.—Problem.To bisect a given rectilineal angle(BAC).

Sol.—InABtake any pointD, and cut off [iii.]AEequal toAD. JoinDE(Post.i.), and upon it, on the side remote fromA, describe the equilateral triangleDEF[i.] JoinAF.AFbisects the given angleBAC.

Dem.—The trianglesDAF,EAFhave the sideADequal toAE(const.) andAFcommon; therefore the two sidesDA,AFare respectively equal toEA,AF, and the baseDFis equal to the baseEF, because they are the sides of an equilateral triangle (Def.xxi.). Therefore [viii.] the angleDAFis equal to the angleEAF;hence the angleBACis bisected by the lineAF.

Cor.—The lineAFis an axis of symmetry of the figure.

Questions for Examination.

1.Why does Euclid describe the equilateral triangle on the side remote fromA?

2.In what case would the construction fail, if the equilateral triangle were described on theother side ofDE?

Exercises.

1.Prove this Proposition without using Prop.viii.

2.Prove thatAFis perpendicular toDE.

3.Prove that any point inAFis equally distant from the pointsDandE.

4.Prove that any point inAFis equally distant from the linesAB,AC.

PROP.X.—Problem.To bisect a given finite right line(AB).

Sol.—UponABdescribe an equilateral triangleACB[i.]. Bisect the angleACBby the lineCD[ix.], meetingABinD,thenABis bisected inD.

Dem.—The two trianglesACD,BCD, have the sideACequal toBC, being the sides of an equilateral triangle, andCDcommon. Therefore the two sidesAC,CDin one are equal to the two sidesBC,CDin the other; and the angleACDis equal to the angleBCD(const.). Therefore the baseADis equal to the baseDB[iv.].HenceABis bisected inD.

Exercises.

1.Show how to bisect a finite right line by describing two circles.

2.Every point equally distant from the pointsA,Bis in the lineCD.

PROP.XI.—Problem.

From a given point(C)in a given right line(AB)to draw a right lineperpendicular to the given line.

Sol.—InACtake any pointD, and makeCEequal toCD[iii.]. UponDEdescribe an equilateral triangleDFE[i.]. JoinCF.ThenCFshall be at right anglestoAB.

Dem.—The two trianglesDCF,ECFhaveCDequal toCE(const.) andCFcommon; therefore the two sidesCD,CFin one are respectively equal to the two sidesCE,CFin the other, and the baseDFis equal to the baseEF, being the sides of an equilateral triangle (Def.xxi.); therefore [viii.] the angleDCEis equal to the angleECF, and they are adjacent angles. Therefore (Def.xiii.) each of them is a right angle,andCFis perpendicular toABat the pointC.

Exercises.

1.The diagonals of a lozenge bisect each other perpendicularly.

2.Prove Prop.xi.without using Prop.viii.

3.Erect a line at right angles to a given line at one of its extremities without producing theline.

4.Find a point in a given line that shall be equally distant from two given points.

5.Find a point in a given line such that, if it be joined to two given points on oppositesides of the line, the angle formed by the joining lines shall be bisected by the givenline.

6.Find a point that shall be equidistant from three given points.

PROP. XII.—Problem.To draw a perpendicular to a given indefinite right line(AB)from a givenpoint(C)without it.

Sol.—Take any pointDon the other side ofAB, and describe (Post.iii.) a circle, withCas centre, andCDas radius, meetingABin the pointsFandG. BisectFGinH[x.]. JoinCH(Post.i.).CHshall be at right angles toAB.

Dem.—JoinCF,CG. Then the two trianglesFHC,GHChaveFHequal toGH(const.), andHCcommon; and the baseCFequal to the baseCG, being radii of the circleFDG(Def.xxxii.). Therefore the angleCHFis equal to the angleCHG[viii.], and, being adjacent angles, they are right angles (Def.xiii.).ThereforeCHis perpendicular toAB.

Exercises.

1.Prove that the circle cannot meetABin more than two points.

2.If one angle of a triangle be equal to the sum of the other two, the triangle can be dividedinto the sum of two isosceles triangles, and the base is equal to twice the line from its middle pointto the opposite angle.

PROP. XIII.—Theorem.

The adjacent angles(ABC,ABD)which one right line(AB)standing on another(CD)makes with it are either both right angles, or their sum is equal to two rightangles.

Dem.—IfABis perpendicular toCD, as in fig. 1, the anglesABC,ABDare right angles. If not, drawBEperpendicular toCD[xi.]. Now the angleCBAis equal to the sum of the two anglesCBE,EBA(Def.xi.). Hence, adding the angleABD, the sum of the anglesCBA,ABDis equal to the sum of the three anglesCBE,EBA,ABD. In like manner, the sum of the anglesCBE,EBDis equal to the sum of the three anglesCBE,EBA,ABD. And things which are equal to the same are equal to one another. Therefore the sum of the anglesCBA,ABDis equal to the sum of the anglesCBE,EBD; butCBE,EBDare right angles;therefore the sum of the anglesCBA,ABDis two rightangles.

Or thus:Denote the angleEBAbyθ; then evidently

the angleCBA= right angle +θ;the angleABD= right angle−θ;thereforeCBA+ABD= two right angles.

Cor.1.—The sum of two supplemental angles is two right angles.

Cor.2.—Two right lines cannot have a common segment.

Cor.3.—The bisector of any angle bisects the corresponding re-entrant angle.

Cor.4.—The bisectors of two supplemental angles are at right angles to each other.

Cor.5.—The angleEBAis half the difference of the anglesCBA,ABD.

PROP. XIV.–Theorem.

If at a point(B)in a right line(BA)two other right lines(CB,BD)onopposite sides make the adjacent angles(CBA,ABD)together equal to two rightangles, these two right lines form one continuous line.

Dem.—IfBDbe not the continuation ofCB, letBEbe its continuation. Now, sinceCBEis a right line, andBAstands on it, the sum of the anglesCBA,ABEis two right angles (xiii.); and the sum of the anglesCBA,ABDis two right angles (hyp.); therefore the sum of the anglesCBA,ABEis equal to the sum of the anglesCBA,ABD. Reject the angleCBA, which is common, and we have the angleABEequal to the angleABD—that is, a part equal to the whole—which is absurd.HenceBDmust be in the same right line withCB.

PROP.XV.—Theorem.If two right lines(AB,CD)intersect one another, the opposite angles areequal(CEA=DEB, andBEC=AED).

Dem.—Because the lineAEstands onCD, the sum of the anglesCEA,AEDis two right angles [xiii.]; and because the lineCEstands onAB, the sum of the anglesBEC,CEAis two right angles; therefore the sum of the anglesCEA,AEDis equal to the sum of the anglesBEC,CEA. Reject the angleCEA, which is common, and we havethe angleAEDequal toBEC. In like manner,the angleCEAis equal toDEB.

The foregoing proof may be briefly given, by saying that opposite angles are equal because they have a common supplement.

Questions for Examination on Props. XIII., XIV., XV.

1.What problem is required in Euclid’s proof of Prop.xiii.?

2.What theorem?Ans.No theorem, only the axioms.

3.If two lines intersect, how many pairs of supplemental angles do they make?

4.What relation does Prop.xiv.bear to Prop.xiii.?

5.What three lines in Prop.xiv.are concurrent?

6.What caution is required in the enunciation of Prop.xiv.?

7.State the converse of Prop.xv. Prove it.

8.What is the subject of Props.xiii.,xiv.,xv.?Ans.Angles at a point.

PROP.XVI.—Theorem.

If any side(BC)of a triangle(ABC)be produced, the exterior angle(ACD)isgreater than either of the interiornon-adjacent angles.

Dem.—BisectACinE[x.]. JoinBE(Post.i.). Produce it, and from the produced part cut offEFequal toBE[iii]. JoinCF. Now becauseECis equal toEA(const.), andEFis equal toEB, the trianglesCEF,AEBhave the sidesCE,EFin one equal to the sidesAE,EBin the other; and the angleCEFequal toAEB[xv.]. Therefore [iv.] the angleECFis equal toEAB; but the angleACDis greater thanECF; therefore the angleACDis greater thanEAB.

In like manner it may be shown, if the sideACbe produced, that the exterior angleBCGis greater than the angleABC; butBCGis equal toACD[xv.]. HenceACDis greater thanABC.ThereforeACDis greater than either of the interiornon-adjacent anglesAorBof the triangleABC.

Cor.1.—The sum of the three interior angles of the triangleBCFis equal to the sum of the three interior angles of the triangleABC.

Cor.2.—The area ofBCFis equal to the area ofABC.

Cor.3.—The linesBAandCF, if produced, cannot meet at any finite distance. For, if they met at any finite pointX, the triangleCAXwould have an exterior angleBACequal to the interior angleACX.

PROP.XVII.—Theorem.Any two angles(B,C)of a triangle(ABC)are together less than two rightangles.

Dem.—ProduceBCtoD; then the exterior angleACDis greater thanABC[xvi.]: to each add the angleACB, and we have the sum of the anglesACD,ACBgreater than the sum of the anglesABC,ACB; but the sum of the anglesACD,ACBis two right angles [xiii.].Therefore the sum of the anglesABC,ACBis lessthan two right angles.

In like manner we may show that the sum of the anglesA,B, or of the anglesA,C, is less than two right angles.

Cor.1.—Every triangle must have at least two acute angles.

Cor.2.—If two angles of a triangle be unequal, the lesser must be acute.

Exercise.

Prove Prop.xvii.without producing a side.

PROP.XVIII.—Theorem.

If in any triangle(ABC)one side(AC)be greater than another(AB), theangle opposite to the greater side is grater than the angle opposite to theless.

Dem.—FromACcut offADequal toAB[iii]. JoinBD(Post.i.). Now sinceABis equal toAD, the triangleABDis isosceles; therefore [v.] the angleADBis equal toABD; but the angleADBis greater than the angleACB[xvi.]; thereforeABDis greater thanACB.Much more is the angleABCgreater than the angleACB.

Or thus:FromAas centre, with the lesser sideABas radius, describe the circleBED, cuttingBCinE. JoinAE. Now sinceABis equal toAE, the angleAEBis equal toABE; butAEBis greater thanACB(xvi.);thereforeABEis greater thanACB.

Exercises.

1.If in the second method the circle cut the lineCBproduced throughB, prove theProposition.

2.This Proposition may be proved by producing the less side.

3.If two of the opposite sides of a quadrilateral be respectively the greatest and least, the anglesadjacent to the least are greater than their opposite angles.

4.In any triangle, the perpendicular from the vertex opposite the side which is not less thaneither of the remaining sides falls within the triangle.

PROP.XIX.—Theorem.

If one angle(B)of a triangle(ABC)be greater than another angle(C), the side(AC)which it opposite to the greater angle is greater than the side(AB)which isopposite to the less.

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