Chapter 15

PROP.XVIII.—TheoremIf a line(CD)touch a circle, the line(OC)from the centre to the point ofcontact is perpendicular to it.

Dem.—If not, suppose another lineOGdrawn from the centre to be perpendicular toCD. LetOGcut the circle inF. Then because the angleOGCis right (hyp.) the angleOCG[I.xvii.] must be acute. Therefore [I.xix.]OCis greater thanOG; butOCis equal toOF[I. Def.xxxii.]; thereforeOFis greater thanOG—that is, a part greater than the whole, which is impossible.HenceOCmust be perpendicular toCD.

Or thus:Since the perpendicular must be the shortest line fromOtoCD, andOCis evidentlythe shortest line; thereforeOCmust be perpendicular toCD.

PROP.XIX.—Theorem.If a line(AB)be a tangent to a circle, the line(AC)drawn at right angles toit from the point of contact passes through the centre.

If the centre be not inAC, letObe the centre. JoinAO. Then becauseABtouches the circle, andOAis drawn from the centre to the point of contact,OAis at right angles toAB[xviii.]; therefore the angleOABis right, and the angleCABis right (hyp.); thereforeOABis equal toCAB—a part equal to the whole, which is impossible.Hence the centre must be in the lineAC.

Cor.—If a number of circles touch the same line at the same point, the locus of their centres is the perpendicular to the line at the point.

Observation.—Propositionsxvi.,xviii.,xix.,are so related that any two can be inferred fromthe third by the “Rule of Identity.” Hence it would, in strict logic, be sufficient to prove any one ofthe three, and the others would follow. Again, these three theorems are limiting cases ofPropositioni.,Cor.1., and Parts 1, 2, of Propositioniii., namely, when the points in which thechord cuts the circle become consecutive.

PROP.XX.—Theorem.The angle(AOB)at the centre(O)of a circle is double the angle(ACB)atthe circumference standing on the same arc.

Dem.—JoinCO, and produce it toE. Then becauseOAis equal toOC, the angleACOis equal toOAC; but the angleAOEis equal to the sum of the two anglesOAC,ACO. Hence the angleAOEis double the angleACO. In like manner the angleEOBis double the angleOCB. Hence (by adding in figs. (α), (β), and subtracting in (γ)),the angleAOBis double of the angleACB.

Cor.—IfAOBbe a straight line,ACBwill be a right angle—that is, the angle ina semicircle is a right angle(comparexxxi.).

PROP.XXI.—Theorem.The angles(ACB,ADB)in the same segment of a circle are equal.

Dem.—LetObe the centre. JoinOA,OB. Then the angleAOBis double of the angleACB[xx.], and also double of the angleADB.Therefore the angleACBisequal to the angleADB.

The following is the proof of the second part—that is, when the arcABis not greater than a semicircle, without using angles greater than two right angles:—

LetObe the centre. JoinCO, and produce it to meet the circle again inE. JoinDE. Now sinceOis the centre, the segmentACEis greater than a semicircle; hence, by the first case, fig. (α), the angleACEis equal toADE. In like manner the angleECBis equal toEDB.Hence the whole angleACBis equal to the whole angleADB.

Cor.1.—If two trianglesACB,ADBon the same baseAB, and on the same side of it, have equal vertical angles, the four pointsA,C,D,Bare concyclic.

Cor.2.—IfA,Bbe two fixed points, and ifCvaries its position in such a way that the angleACBretains the same value throughout, the locus ofCis a circle.

In other words—Given the base of a triangle and the vertical angle, the locus ofthe vertex is a circle.

Exercises.

1.Given the base of a triangle and the vertical angle, find the locus—

(1) of the intersection of its perpendiculars;

(2) of the intersection of the internal bisectors of its base angles;

(3) of the intersection of the external bisectors of the base angles;

(4) of the intersection of the external bisector of one base angle and the internal bisector ofthe other.

2.If the sum of the squares of two lines be given, their sum is a maximum when the lines areequal.

3.Of all triangles having the same base and vertical angle, the sum of the sides of an isoscelestriangle is a maximum.

4.Of all triangles inscribed in a circle, the equilateral triangle has the maximumperimeter.

5.Of all concyclic figures having a given number of sides, the area is a maximum when the sidesare equal.

PROP.XXII.—Theorem.The sum of the opposite angles of a quadrilateral(ABCD)inscribed in a circleis two right angles.

Dem.—JoinAC,BD. The angleABDis equal toACD, being in the same segmentABCD[xxi.]; and the angleDBCis equal toDAC, because they are in the same segmentDABC. Hence the whole angleABCis equal to the sum of the two anglesACD,DAC. To each add the angleCDA, and we have the sum of the two anglesABC,CDAequal to the sum of the three anglesACD,DAC,CDAof the triangleACD; but the sum of the three angles of a triangle is equal to two right angles [I.xxxii.].Therefore the sum ofABC,CDAis two rightangles.

Or thus:LetObe the centre of the circle. JoinOA,OC(seefig. 2). Now the angleAOCis double ofCDA[xx.], and the angleCOAis double ofABC. Hence the sum of the angles [I. Def.ix., note]AOC,COAis double of the sum of the anglesCDA,ABC; but the sum of two anglesAOC,COAis four right angles.Therefore the sum of the anglesCDA,ABCis two rightangles.

Or again:LetObe the centre (fig. 2). JoinOA,OB OC,OD. Then the four trianglesAOB,BOC,COD,DOAare each isosceles. Hence the angleOABis equal to the angleOBA, and the angleOADequal to the angleODA; therefore the angleBADis equal to the sum of the anglesOBA,ODA. In like manner the angleBCDis equal to the sum of the anglesOBC,ODC. Hence the sum of the two anglesBAD,BCDis equal to the sum of the two anglesABC,ADC,and hence each sum is tworight angles.

Cor.—If a parallelogram be inscribed in a circle it is a rectangle.

Exercises.

1.If the opposite angles of a quadrilateral be supplemental, it is cyclic.

2.If a figure of six sides be inscribed in a circle, the sum of any three alternate angles is fourright angles.

3.A line which makes equal angles with one pair of opposite sides of a cyclic quadrilateral,makes equal angles with the remaining pair and with the diagonals.

4.If two opposite sides of a cyclic quadrilateral be produced to meet, and a perpendicular be letfall on the bisector of the angle between them from the point of intersection of the diagonals, thisperpendicular will bisect the angle between the diagonals.

5.If two pairs of opposite sides of a cyclic hexagon be respectively parallel to each other, theremaining pair of sides are also parallel.

6.If two circles intersect in the pointsA,B, and any two linesACD,BFE, be drawn throughAandB, cutting one of the circles in the pointsC,E, and the other in the pointsD,F, the lineCEis parallel toDF.

7.If equilateral triangles be described on the sides of any triangle, the lines joining thevertices of the original triangle to the opposite vertices of the equilateral triangles areconcurrent.

8.In the same case prove that the centres of the circles described about the equilateral trianglesform another equilateral triangle.

9.If a quadrilateral be described about a circle, the angles at the centre subtended by theopposite sides are supplemental.

10.The perpendiculars of a triangle are concurrent.

11.If a variable tangent meets two parallel tangents it subtends a right angle at thecentre.

12.The feet of the perpendiculars let fall on the sides of a triangle from any point in thecircumference of the circumscribed circle are collinear (Simson).

Def.—The line of collinearity is called Simson’s line.

13.If a hexagon be circumscribed about a circle, the sum of the angles subtended at the centreby any three alternate sides is equal to two right angles.

PROP.XXIII—Theorem.Two similar segments of circles which do not coincide cannot be constructedon the same chord(AB), and on the same side of that chord.

Dem.—If possible, letACB,ADB, be two similar segments constructed on the same side ofAB. Take any pointDin the inner one. JoinAD, and produce it to meet the outer one inC. JoinBC,BD. Then since the segments are similar, the angleADBis equal toACB(Def.x.), which is impossible [I.xvi.].Hence twosimilar segments not coinciding cannot be described on the same chord and on thesame side of it.

PROP.XXIV.—Theorem.Similar segments of circles(AEB,CFD)on equal chords(AB,CD)areequal to one another.

Dem.—Since the lines are equal, ifABbe applied toCD, so that the pointAwill coincide withC, and the lineABwithCD, the pointBshall coincide withD; and because the segments are similar, they must coincide [xxiii.].Hence they areequal.

This demonstration may be stated as follows:—Since the chords are equal, they are congruent;and therefore the segments, being similar, must be congruent.

PROP.XXV.—Problem.An arc(ABC)of a circle being given, it is required to describe the whole circle.

Sol.—Take any three pointsA,B,Cin the arc. JoinAB,BC. BisectABinD, andBCinE. ErectDF,EFat right angles toAB,BC; thenF, the point of intersection, will be the centre of the circle.

Dem.—BecauseDFbisects the chordABand is perpendicular to it, it passes through the centre [i.,Cor.1]. In like mannerEFpasses through the centre.Hencethe pointFmust be the centre; and the circle described fromFas centre, withFAasradius, will be the circle required.

PROP.XXVI.—Theorem.

The four Propositionsxxvi.–xxix.are so like in their enunciations that studentsfrequently substitute one for another. The following scheme will assist in rememberingthem:—

In Propositionxxvi.aregivenangles=, toprovearcs=,,,xxvii.,,arcs=,,,angles=,,,xxviii.,,chords=,,,arcs=,,,xxix.,,arcs=,,,chords=;

so that Propositionxxvii.is the converse ofxxvi., andxxix.ofxxviii.

In equal circles(ACB,DFE), equal angles at the centres(AOB,DHE)orat the circumferences(ACB,DFE)stand upon equal arcs.

Dem.—1. Suppose the angles at the centres to be given equal. Now because the circles are equal their radii are equal (Def.i.). Therefore the two trianglesAOB,DHEhave the sidesAO,OBin one respectively equal to the sidesDH,HEin the other, and the angleAOBequal toDHE(hyp.). Therefore [I.iv.] the baseABis equal toDE.

Again, since the anglesACB,DFEare [xx.] the halves of the equal anglesAOB,DHE, they are equal [I. Axiomvii.]. Therefore (Def.x.) the segmentsACB,DFEare similar, and their chordsAB,DEhave been proved equal; therefore [xxiv.] the segments are equal. And taking these equals from the whole circles, which are equal (hyp.), the remaining segmentsAGB,DKEare equal.Hence the arcsAGB,DKEare equal.

2. The demonstration of this case is included in the foregoing.

Cor.1.—If the opposite angles of a cyclic quadrilateral be equal, one of its diagonals must be a diameter of the circumscribed circle.

Cor.2.—Parallel chords in a circle intercept equal arcs.

Cor.3.—If two chords intersect at any point within a circle, the sum of the opposite arcs which they intercept is equal to the arc which parallel chords intersecting on the circumference intercept. 2. If they intersect without the circle, the difference of the arcs they intercept is equal to the arc which parallel chords intersecting on the circumference intercept.

Cor.4.—If two chords intersect at right angles, the sum of the opposite arcs which they intercept on the circle is a semicircle.

PROP.XXVII.—Theorem.

In equal circles(ACB,DFE), angles at the centres(AOB,DHE), or at thecircumferences(ACB,DFE), which stand on equal arcs(AB,DE), areequal.

Dem.—If possible let one of them, such asAOB, be greater than the other,DHE; and suppose a part such asAOLto be equal toDHE. Then since the circles are equal, and the anglesAOL,DHEat the centres are equal (hyp.), the arcALis equal toDE[xxvi.]; butABis equal toDE(hyp.). HenceALis equal toAB—that is, a part equal to the whole, which is absurd.Therefore the angleAOBis equal toDHE.

2.The angles at the circumference, being the halves of the central angles, aretherefore equal.

PROP.XXVIII.—Theorem.

In equal circles(ACB,DFE), equal chords(AB,DE)divide the circumferencesinto arcs, which are equal each to each—that is, the lesser to the lesser, and thegreater to the greater.

Dem.—If the equal chords be diameters, the Proposition is evident. If not, letO,Hbe the centres. JoinAO,OB,DH,HE; then because the circles are equal their radii are equal (Def.i.). Hence the two trianglesAOB,DHEhave the sidesAO,OBin one respectively equal to the sidesDH,HEin the other, and the baseABis equal toDE(hyp.). Therefore [I.viii.] the angleAOBis equal toDHE.Hence the arcAGBis equal toDKE[xxvi.]; and since the whole circumferenceAGBCis equal to the whole circumferenceDKEF,the remaining arcACBis equal to the remaining arcDFE.

Exercises.

1.The line joining the feet of perpendiculars from any point in the circumference of a circle, ontwo diameters given in position, is given in magnitude.

2.If a line of given length slide between two lines given in position, the locus of the intersectionof perpendiculars to the given lines at its extremities is a circle. (This is the converse of1.)

PROP.XXIX.—Theorem.In equal circles(ACB,DFE), equal arcs(AGB,DCK)are subtended byequal chords.

Dem.—LetO,Hbe the centres (seelast fig.). JoinAO,OB,DH,HE; then because the circles are equal, the anglesAOB,DHEat the centres, which stand on the equal arcsAGB,DKE, are equal [xxvii.]. Again, because the trianglesAOB,DHEhave the two sidesAO,OBin one respectively equal to the two sidesDH,HEin the other, and the angleAOBequal to the angleDHE,the baseABof one isequal to the baseDEof the other.

Observation.—Since the two circles in the four last Propositions are equal, they are congruentfigures, and the truth of the Propositions is evident by superposition.

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