Chapter 16

PROP.XXX.—Problem.To bisect a given arcACB.

Sol.—Draw the chordAB; bisect it inD; erectDCat right angles toAB, meeting the arc inC; then the arc is bisected inC.

Dem.—JoinAC,BC. Then the trianglesADC,BDChave the sideADequal toDB(const.), andDCcommon to both, and the angleADCequal to the angleBDC, each being right. Hence the baseACis equal to the baseBC. Therefore [xxviii.] the arcACis equal to the arcBC.Hence the arcABis bisected inC.

Exercises.

1.ABCDis a semicircle whose diameter isAD; the chordBCproduced meetsADproduced inE: prove that ifCEis equal to the radius, the arcABis equal to three timesCD.

2.The internal and the external bisectors of the vertical angle of a triangle inscribed in acircle meet the circumference again in points equidistant from the extremities of thebase.

3.If fromA, one of the points of intersection of two given circles, two chordsACD,AC′D′be drawn, cutting the circles in the pointsC,D;C′,D′, the trianglesBCD,BC′D′, formed by joining these to the second pointBof intersection of the circles, areequiangular.

4.If the vertical angleACBof a triangle inscribed in a circle be bisected by a lineCD, whichmeets the circle again inD, and fromDperpendicularsDE,DFbe drawn to the sides, one of whichmust be produced: prove thatEAis equal toBF, and hence show thatCEis equal to half the sumofAC,BC.

PROP.XXXI.—Theorem.

In a circle—(1). The angle in a semicircle is a right angle.(2). The angle in asegment greater than a semicircle is an acute angle.(3). The angle in a segment lessthan a semicircle is an obtuse angle.

Dem.—(1). LetABbe the diameter,Cany point in the semicircle. JoinAC,CB.The angleACBis a right angle.

For letObe the centre. JoinOC, and produceACtoF. Then becauseAOis equal toOC, the angleACOis equal to the angleOAC. In like manner, the angleOCBis equal toCBO. Hence the angleACBis equal to the sum of the two anglesBAC,CBA; but [I.xxxii.] the angleFCBis equal to the sum of the two interior anglesBAC,CBAof the triangleABC. Hence the angleACBis equal to its adjacent angleFCB,and therefore it is a right angle[I. Def.xiii.].

(2). Let the arcACEbe greater than a semicircle. JoinCE. Then the angleACEis evidently less thanACB; butACBis right;thereforeACEisacute.

(3). Let the arcACDbe less than a semicircle; then evidently, from (1),the angleACDis obtuse.

Cor.1.—If a parallelogram be inscribed in a circle, its diagonals intersect at the centre of the circle.

Cor.2.—Find the centre of a circle by means of a carpenter’s square.

Cor.3.—From a point outside a circle draw two tangents to the circle.

PROP.XXXII.—Theorem.

If a line(EF)be a tangent to a circle, and from the point of contact(A)achord(AC)be drawn cutting the circle, the angles made by this line with thetangent are respectively equal to the angles in the alternate segments of thecircle.

Dem.—(1). If the chord passes through the centre, the Proposition is evident, for the angles are right angles; but if not, from the point of contactAdrawABat right angles to the tangent. JoinBC. Then becauseEFis a tangent to the circle, andABis drawn from the point of contact perpendicular toEF,ABpasses through this centre [xix.]. Therefore the angleACBis right [xxxi.]. Hence the sum of the two remaining anglesABC,CABis one right angle; but the angleBAFis right (const.); therefore the sum of the anglesABC,BACis equal toBAF. RejectBAC, which is common, and we getthe angleABCequal to the angleFAC.

(2). Take any pointDin the arcAC. It is required to prove that the angleCAEis equal toCDA.

Since the quadrilateralABCDis cyclic, the sum of the opposite anglesABC,CDAis two right angles [xxii.], and therefore equal to the sum of the anglesFAC,CAE; but the anglesABC,FACare equal (1). Reject them, and we get theangleCDAequal toCAE.

Or thus:Take any pointGin the semicircleAGB. JoinAG,GB,GC. Then the angleAGB=FAB, each being right, andCGB=CAB[xxi.]. Therefore the remaining angleAGC=FAC. Again, joinBD,CD. The angleBDA=BAE, each being right, andCDB=CAB[xxi.].Hence the angleCDA=CAE.—Lardner.

Or by the method of limits, seeTownsend’sModern Geometry, vol. i., page 14.

The angleBACis equal toBDC[xxi.]. Now let the pointBmove until it becomes consecutive toA; thenABwill be a tangent, andBDwill coincide withAD, and the angleBDCwithADC. Hence, ifAXbe a tangent atA,ACany chord,the angle which the tangent makes with the chord is equal to the angle in the alternatesegment.

Exercises.

1.If two circles touch, any line drawn through the point of contact will cut off similarsegments.

2.If two circles touch, and any two lines be drawn through the point of contact, cutting bothcircles again, the chord connecting their points of intersection with one circle is parallel to the chordconnecting their points of intersection with the other circle.

3.ACBis an arc of a circle,CEa tangent atC, meeting the chordABproduced inE, andADa perpendicular toABinD: prove, ifDEbe bisected inC, that the arcAC= 2CB.

4.If two circles touch at a pointA, andABCbe a chord throughA, meeting the circles inBandC: prove that the tangents atBandCare parallel to each other, and that when one circle iswithin the other, the tangent atBmeets the outer circle in two points equidistant fromC.

5.If two circles touch externally, their common tangent at either side subtends a rightangle at the point of contact, and its square is equal to the rectangle contained by theirdiameters.

PROP.XXXIII.—Problem.On a given right line(AB)to describe a segment of a circle which shallcontain an angle equal to a given rectilineal angle(X).

Sol.—IfXbe a right angle, describe a semicircle on the given line, and the thing required is done; for the angle in a semicircle is a right angle.

If not, make with the given lineABthe angleBAEequal toX. ErectACat right angles toAE, andBCat right angles toAB. OnACas diameter describe a circle:it will be the circle required.

Dem.—The circle onACas diameter passes throughB, since the angleABCis right [xxxi.] and touchesAE, since the angleCAEis right [xvi.]. Therefore the angleBAE[xxxii.] is equal to the angle in the alternate segment; but the angleBAEis equal to the angleX(const.).Therefore the angleXis equal to the angle inthe segment described onAB.

Exercises.

1.Construct a triangle, being given base, vertical angle, and any of the following data:—1.Perpendicular. 2. The sum or difference of the sides. 3. Sum or difference of the squares ofthe sides. 4. Side of the inscribed square on the base. 5. The median that bisects thebase.

2.If lines be drawn from a fixed point to all the points of the circumference of a given circle, thelocus of all their points of bisection is a circle.

3.Given the base and vertical angle of a triangle, find the locus of the middle point of the linejoining the vertices of equilateral triangles described on the sides.

4.In the same case, find the loci of the angular points of a square described on one of thesides.

PROP.XXXIV.—Problem.To cut off from a given circle(ABC)a segment which shall contain an angleequal to a given angle(X).

Sol.—Take any pointAin the circumference. Draw the tangentAD, and make the angleDACequal to the given angleX.ACwill cut off the required segment.

Dem.—Take any pointBin the alternate segment. JoinBA,BC. Then the angleDACis equal toABC[xxxii.]; butDACis equal toX(const.).Therefore the angleABCis equal toX.

PROP.XXXV.—Theorem.If two chords(AB,CD)of a circle intersect in a point(E)within the circle,the rectangles(AE.EB,CE.ED)contained by the segments are equal.

Dem.—1. If the point of intersection be the centre, each rectangle is equal to the square of the radius.Hence they are equal.

2. Let one of the chordsABpass through the centreO, and cut the other chordCD, which does not pass through the centre, at right angles. JoinOC. Now becauseABpasses through the centre, and cuts the other chordCD, which does not pass through the centre at right angles, it bisects it [iii.]. Again, becauseABis divided equally inOand unequally inE, the rectangleAE.EB, together withOE2, is equal toOB2—that is, toOC2[II.v.]; butOC2is equal toOE2+EC2[I.xlvii.] Therefore

2 2 2 AE.EB + OE = OE + EC .

RejectOE2, which is common, and we haveAE.EB=EC2; butCE2is equal to the rectangleCE.ED, sinceCEis equal toED.Therefore the rectangleAE.EBisequal to the rectangleCE.ED.

3. LetABpass through the centre, and cutCD, which does not pass through the centre obliquely. LetObe the centre. DrawOFperpendicular toCD[I.xi.]. JoinOC,OD. Then, sinceCDis cut at right angles byOF, which passes through the centre, it is bisected inF[iii.], and divided unequally inE. Hence

Hence, adding, sinceFE2+OF2=OE2[I.xlvii.], andFD2+OF2=OD2, we get

CE.ED + OE2 = OD or OB2.

Again, sinceABis bisected inOand divided unequally inE,

4. Let neither chord pass through the centre. Through the pointE, where they intersect, draw the diameterFG. Then by 3, the rectangleFE.EGis equal to the rectangleAE.EB, and also to the rectangleCE.ED.Hence the rectangleAE.EBisequal to the rectangleCE.ED.

Cor.1.—If a chord of a circle be divided in any point within the circle, the rectangle contained by its segments is equal to the difference between the square of the radius and the square of the line drawn from the centre to the point of section.

Cor.2.—If the rectangle contained by the segments of one of two intersecting lines be equal to the rectangle contained by the segments of the other, the four extremities are concyclic.

Cor.3.—If two triangles be equiangular, the rectangle contained by thenon-corresponding sides about any two equal angles are equal.

LetABO,DCObe the equiangular triangles, and let them be placed so that the equal angles atOmay be vertically opposite, and that the non-corresponding sidesAO,COmay be in one line; then the non-corresponding sidesBO,ODshall be in one line. Now, since the angleABDis equal toACD, the pointsA,B,C,Dare concyclic [xxi.,Cor.1].Hence the rectangleAO.OCis equal to the rectangleBO.OD[xxxv.].

Exercises.

1.In any triangle, the rectangle contained by two sides is equal to the rectanglecontained by the perpendicular on the third side and the diameter of the circumscribedcircle.

Def.—The supplement of an arc is the difference between it and a semicircle.

2.The rectangle contained by the chord of an arc and the chord of its supplement is equal tothe rectangle contained by the radius and the chord of twice the supplement.

3.If the base of a triangle be given, and the sum of the sides, the rectangle contained by theperpendiculars from the extremities of the base on the external bisector of the vertical angle isgiven.

4.If the base and the difference of the sides be given, the rectangle contained by theperpendiculars from the extremities of the base on the internal bisector is given.

5.Through one of the points of intersection of two circles draw a secant, so that the rectanglecontained by the intercepted chords may be given, or a maximum.

6.If the sum of two arcs,AC,CBof a circle be less than a semicircle, the rectangleAC.CBcontained by their chords is equal to the rectangle contained by the radius, and the excess of thechord of the supplement of their difference above the chord of the supplement of theirsum.—Catalan.

Dem.—DrawDE, the diameter which is perpendicular toAB, and draw the chordsCF,BGparallel toDE. Now it is evident that the difference between the arcsAC,CBis equal to 2CD, andtherefore =CD+EF. Hence the arcCBFis the supplement of the difference, andCFis the chordof that supplement. Again, since the angleABGis right, the arcABGis a semicircle. HenceBGisthe supplement of the sum of the arcsAC,CB; therefore the lineBGis the chord of the supplementof the sum. Now (Ex.1), the rectangleAC.CBis equal to the rectangle contained by the diameterandCI, and therefore equal to the rectangle contained by the radius and 2CI; but thedifference betweenCFandBGis evidently equal to 2CI.Hence the rectangleAC.CBisequal to the rectangle contained by the radius and the difference between the chordsCF,BG.

7.If we joinAF,BFwe find, as before, the rectangleAF.FBequal to the rectangle containedby the radius and 2FI—that is, equal to the rectangle contained by the radius and the sum ofCFandBG. Hence—If the sum of two arcs of a circle be greater than a semicircle, the rectanglecontained by their chords is equal to the rectangle contained by the radius, and the sum of thechords of the supplements of their sum and their difference.

8.Through a given point draw a transversal cutting two lines given in position, sothat the rectangle contained by the segments intercepted between it and the line may begiven.

PROP.XXXVI.—Theorem.

If from any point(P)without a circle two lines be drawn to it, one of which(PT)isa tangent, and the other(PA)a secant, the rectangle(AP,BP)contained by thesegments of the secant is equal to the square of the tangent.

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