Exercises.
Exercises.
1.In the figure [I.xlvii.] ifEF,GKbe joined, proveEF2−CO2= (AB+BO)2.
2.ProveGK2−EF2= 3AB(AO−BO).
3.11Ex.3 occurs in the solution of the problem of the inscription of a regular polygon of seventeensides in a circle.Seenote C.Given the difference of two lines =R, and their rectangle = 4R2; find the lines.
PROP.IX.—Theorem.
PROP.IX.—Theorem.
If a line(AB)be bisected(atC)and divided into two unequal parts(atD), the sumof the squares on the unequal parts(AD,DB)is double the sum of thesquares on half the line(AC), and on the segment(CD)between the points ofsection.
Dem.—ErectCEat right angles toAB, and make it equal toACorCB. JoinAE,EB. DrawDFparallel toCE, andFGparallel toCD. JoinAF.
BecauseACis equal toCE, and the angleACEis right, the angleCEAis half a right angle. In like manner the anglesCEB,CBEare half right angles; therefore the whole angleAEFis right. Again, becauseGFis parallel toCB, andCEintersects them, the angleEGFis equal toECB; butECBis right (const.); thereforeEGFis right; andGEFhas been proved to be half a right angle; therefore the angleGFEis half a right angle [I.xxxii.]. Therefore [I.vi.]GEis equal toGF. In like mannerFDis equal toDB.
Again, sinceACis equal toCE,AC2is equal toCE2; butAE2is equal toAC2+CE2[I.xlvii.]. ThereforeAE2is equal to 2AC2. In like mannerEF2is equal to 2GF2or 2CD2. ThereforeAE2+EF2is equal to 2AC2+ 2CD2; butAE2+EF2is equal toAF2[I.xlvii.]. ThereforeAF2is equal to 2AC2+ 2CD2.
Again, sinceDFis equal toDB,DF2is equal toDB2: to each addAD2, and we getAD2+DF2equal toAD2+DB2; butAD2+DF2is equal toAF2; thereforeAF2is equal toAD2+DB2; and we have provedAF2equal to 2AC2+ 2CD2.ThereforeAD2+DB2is equal to2AC2+ 2CD2.
Exercises.
Exercises.
1.The sum of the squares on the segments of a line of given length is a minimum when it isbisected.
2.Divide a given line internally, so that the sum of the squares on the parts may be equal to agiven square, and state the limitation to its possibility.
3.If a lineABbe bisected inCand divided unequally inD,
AD2 + DB2 =2AD.DB + 4CD2.
4.Twice the square on the line joining any point in the hypotenuse of a right-angledisosceles triangle to the vertex is equal to the sum of the squares on the segments of thehypotenuse.
5.If a line be divided into any number of parts, the continued product of all theparts is a maximum, and the sum of their squares is a minimum when all the parts areequal.
PROP.X.—Theorem.
PROP.X.—Theorem.
If a line(AB)be bisected(atC)and divided externally(atD), the sum of thesquares on the segments(AD,DB)made by the external point is equal to twice thesquare on half the line, and twice the square on the segment between the points ofsection.
Dem.—ErectCEat right angles toAB, and make it equal toACorCB. JoinAE,EB. DrawDFparallel toCE, and produceEB. Now sinceDFis parallel toEC, the angleBDFis = toBCE[I.xxix.], and [I.xv.] the angleDBFis = toEBC; but the sum of the anglesBCE,EBCis less than two right angles [I.xvii.]; therefore the sum of the anglesBDF,DBFis less than two right angles, and therefore [I., Axiomxii.] the linesEB,DF, if produced, will meet. Let them meet inF. ThroughFdrawFGparallel toAB, and produceECto meet it inG. JoinAF.
BecauseACis equal toCE, and the angleACEis right, the angleCEAis half a right angle. In like manner the anglesCEB,CBEare half right angles; therefore the whole angleAEFis right. Again, becauseGFis parallel toCB, andGEintersects them, the angleEGFis equal toECB[I.xxix.]; butECBis right (const.); thereforeEGFis right, andGEFhas been proved to be half a right angle; therefore [I.xxxii.]GFEis half a right angle, and therefore [I.vi.]GEis equal toGF. In like mannerFDis equal toDB.
Again, sinceACis equal toCE,AC2is equal toCE2; butAE2is equal toAC2+CE2[I.xlvii.]; thereforeAE2is equal to 2AC2. In like mannerEF2is equal to 2GF2or 2CD2; thereforeAE2+EF2is equal to 2AC2+ 2CD2; butAE2+EF2is equal toAF2[I.xlvii.]. ThereforeAF2is equal to 2AC2+ 2CD2.
Again, sinceDFis equal toDB,DF2is equal toDB2: to each addAD2, and we getAD2+DF2equal toAD2+DB2; butAD2+DF2is equal toAF2; thereforeAF2is equal toAD2+DB2; andAF2has been proved equal to 2AC2+ 2CD2.ThereforeAD2+DB2is equal to2AC2+ 2CD2.
Square and add, and we getAD2+BD2= 2CD2+ 2AC2.
The following enunciations include Propositionsix. andx.:—
1.The square on the sum of any two lines plus the square on their differenceequal twice the sum of their squares.
2.The sum of the squares on any two lines it equal to twice the square on half thesum plus twice the square on half the difference of the lines.
3.If a line be cut into two unequal parts, and also into two equal parts, the sum ofthe squares on the two unequal parts exceeds the sum of the squares on the two equalparts by the sum of the squares of the two differences between the equal and unequalparts.
Exercises
Exercises
.
1.Given the sum or the difference of any two lines, and the sum of their squares; find thelines.
2.The sum of the squares on two sidesAC,CBof a triangle is equal to twice the square on halfthe baseAB, and twice the square on the median which bisectsAB.
3.If the base of a triangle be given both in magnitude and position, and the sum of the squareson the sides in magnitude, the locus of the vertex is a circle.
4.If in the△ABCa pointDin the baseBCbe such that
BA2+ BD2 =CA2 + CD2;
prove that the middle point ofADis equally distant fromBandC.
5.The sum of the squares on the sides of a parallelogram is equal to the sum of the squares onits diagonals.
PROP.XI.—Problem.
PROP.XI.—Problem.
To divide a given finite line(AB)into two segments(inH), so that the rectangle(AB.BH)contained by the whole line and one segment may be equal to the square onthe other segment.
Sol.—OnABdescribe the squareABDC[I.xlvi.]. BisectACinE. JoinBE. ProduceEAtoF, and makeEFequal toEB. OnAFdescribe the squareAFGH.His the point required.
Dem.—ProduceGHtoK. Then becauseCAis bisected inE, and divided externally inF, the rectangleCF.AF, together with the square onEA, is equal to the square onEF[vi.]; butEFis equal toEB(const.); therefore the rectangleCF.AF, together withEA2, is equal toEB2—that is [I.xlvii.] equal toEA2+AB2. RejectingEA2, which is common, we get the rectangleCF.AFequal toAB2. Again, sinceAFis equal toFG, being the sides of a square, the rectangleCF.AFis equal toCF.FG—that is, to the figureCG; andAB2is equal to the figureAD; thereforeCGis equal toAD. Reject the partAK, which is common, and we get the figureFHequal to the figureHD; butHDis equal to the rectangleAB.BH, becauseBDis equal toAB, andFHis the square onAH.Therefore the rectangleAB.BHis equal to the square onAH.
Def.—A line divided as in this Proposition is said to be divided in “extreme andmean ratio.”
Cor.1.—The line CFis divided in “extreme and mean ratio” atA.
Cor.2.—If from the greater segmentCAofCFwe take a segment equal toAF, it is evident thatCAwill be divided into parts respectively equal toAH,HB. Hence, if a line be divided in extreme and mean ratio, the greater segment will be cut in the same manner by taking on it a part equal to the less; and the less will be similarly divided by taking on it a part equal to the difference, and so on, &c.
Cor.3.—LetABbe divided in “extreme and mean ratio” inC, then it is evident (Cor.2) thatACis greater thanCB. Cut offCD=CB; then (Cor.2)ACis cut in “extreme and mean ratio” atD, andCDis greater thanAD. Next, cut offDEequal toAD, and in the same manner we haveDEgreater thanEC, and so on. Now sinceCDis greater thanAD, it is evident thatCDis not a common measure ofACandCB, and therefore not a common measure ofABandAC. In like mannerADis not a common measure ofACandCD, and therefore not a common measure ofABandAC. Hence, no matter how far we proceed we cannot arrive at any remainder which will be a common measure ofABandAC.Hence, the parts of a line divided in “extreme and mean ratio”areincommensurable.
Exercises.
Exercises.
1.Cut a line externally in “extreme and mean ratio.”
2.The difference between the squares on the segments of a line divided in “extreme and meanratio” is equal to their rectangle.
3.In a right-angled triangle, if the square on one side be equal to the rectangle contained by thehypotenuse and the other side, the hypotenuse is cut in “extreme and mean ratio” by theperpendicular on it from the right angle.
4.IfABbe cut in “extreme and mean ratio” atC, prove that
5.The three lines joining the pairs of pointsG,B;F,D;A,K, in the construction ofPropositionxi., are parallel.
6.IfCHintersectBEinO,AOis perpendicular toCH.
7.IfCHbe produced, it meetsBFat right angles.
8.ABCis a right-angled triangle havingAB= 2AC: ifAHbe made equal to the differencebetweenBCandAC,ABis divided in “extreme and mean ratio” atH.
PROP.XII.—Theorem.
PROP.XII.—Theorem.
In an obtuse-angled triangle(ABC), the square on the side(AB)subtending theobtuse angle exceeds the sum of the squares on the sides(BC,CA)containing theobtuse angle, by twice the rectangle contained by either of them(BC), and itscontinuation(CD)to meet a perpendicular(AD)on it from the oppositeangle.
Dem.—BecauseBDis divided into two parts inC, we have
Hence, adding, since [I.xlvii.]BD2+AD2=AB2, andCD2+AD2=CA2, we get
2 2 2 AB = BC + CA + 2BC.CD.
ThereforeAB2is greater thanBC2+CA2by2BC.CD.
The foregoing proof differs from Euclid’s only in the use of symbols. I have found by experiencethat pupils more readily understand it than any other method.
Or thus:By the First Book: Describe squares on the three sides. DrawAE,BF,CGperpendicular to the sides of the squares. Then it can be proved exactly as in the demonstration of[I.xlvii.], that the rectangleBGis equal toBE,AGtoAF, andCEtoCF.Hence the sum of thetwo squares onAC,CBis less than the square onABby twice the rectangleCE; that is, by twicethe rectangleBC.CD.
Cor.1.—If perpendiculars fromAandBto the opposite sides meet them inHandD, therectangleAC.CHis equal to the rectangleBC.CD.
Exercises.
Exercises.
1.If the angleACBof a triangle be equal to twice the angle of an equilateral triangle,AB2=BC2+CA2+BC.CA.
2.ABCDis a quadrilateral whose opposite anglesBandDare right, andAD,BCproducedmeet inE; proveAE.DE=BE.CE.
3.ABCis a right-angled triangle, andBDis a perpendicular on the hypotenuseAC; ProveAB.DC=BD.BC.
4.If a lineABbe divided inCso thatAC2= 2CB2; prove thatAB2+BC2= 2AB.AC.
5.IfABbe the diameter of a semicircle, find a pointCinABsuch that, joiningCto a fixedpointDin the circumference, and erecting a perpendicularCEmeeting the circumference inE,CE2−CD2may be equal to a given square.
6.If the square of a lineCD, drawn from the angleCof an equilateral triangleABCto a pointDin the sideABproduced, be equal to 2AB2; prove thatADis cut in “extreme and mean ratio” atB.