Cor.2.—The square on a line is equal to four times the square on itshalf.
For letAB= 2AC, thenAB2= 4AC2.
ThisCor.may be proved by the First Book thus: ErectCDat right angles toAB, and makeCD=ACorCB. JoinAD,DB.
But since the angleADBis right,AD2+DB2=AB2;
Cor.3.—If a line be divided into any number of parts, the square onthe whole is equal to the sum of the squares on all the parts, together withtwice the sum of the rectangles contained by the several distinct pairs ofparts.
Exercises.
Exercises.
1.Prove Propositioniv.by using Propositionsii.andiii.
2.If from the vertical angle of a right-angled triangle a perpendicular be let fall onthe hypotenuse, its square is equal to the rectangle contained by the segments of thehypotenuse.
3.From the hypotenuse of a right-angled triangle portions are cut off equal to the adjacentsides; prove that the square on the middle segment is equal to twice the rectangle contained by theextreme segments.
4.In any right-angled triangle the square on the sum of the hypotenuse and perpendicular, fromthe right angle on the hypotenuse, exceeds the square on the sum of the sides by the square on theperpendicular.
5.The square on the perimeter of a right-angled triangle is equal to twice the rectanglecontained by the sum of the hypotenuse and one side, and the sum of the hypotenuse and the otherside.
PROP.V.—Theorem.
PROP.V.—Theorem.
If a line(AB)be divided into two equal parts(atC), and also into two unequal parts(atD), the rectangle(AD.DB)contained by the unequal parts, together with thesquare on the part(CD)between the points of section, is equal to the square on halfthe line.
Dem.—OnCBdescribe the squareCBEF[I.xlvi.]. JoinBF. ThroughDdrawDGparallel toCF, meetingBFinH. ThroughHdrawKMparallel toAB, and throughAdrawAKparallel toCL[I.xxxi.].
The parallelogramCMis equal toDE[I.xliii.,Cor.2]; butALis equal toCM[I.xxxvi.], because they are on equal basesAC,CB, and between the same parallels; thereforeALis equal toDE: to each addCH, and we get the parallelogramAHequal to the gnomonCMG; butAHis equal to the rectangleAD.DH, and therefore equal to the rectangleAD.DB, sinceDHis equal toDB[iv.,Cor.1]; therefore the rectangleAD.DBis equal to the gnomonCMG, and the square onCDis equal to the figureLG.Hence the rectangleAD.DB, together withthe square onCD, is equal to the whole figureCBEF—that is, to the square onCB.
Cor. 1.—The rectangleAD.DBis the rectangle contained by the sum of the linesAC,CDand their difference; and we have proved it equal to the difference between the square onACand the square onCD.Hence the difference of thesquares on two lines is equal to the rectangle contained by their sum and theirdifference.
Cor. 2.—The perimeter of the rectangleAHis equal to 2AB, and is therefore independent of the position of the pointDon the lineAB; and the area of the same rectangle is less than the square on half the line by the square on the segment betweenDand the middle point of the line; therefore, whenDis the middle point, the rectangle will have the maximum area.Hence, of all rectangles having the same perimeter, the square has the greatestarea.
Exercises.
Exercises.
1.Divide a given line so that the rectangle contained by its parts may have a maximumarea.
2.Divide a given line so that the rectangle contained by its segments may be equal to a givensquare, not exceeding the square on half the given line.
3.The rectangle contained by the sum and the difference of two sides of a triangle is equal tothe rectangle contained by the base and the difference of the segments of the base, made by theperpendicular from the vertex.
4.The difference of the sides of a triangle is less than the difference of the segments of the base,made by the perpendicular from the vertex.
5.The difference between the square on one of the equal sides of an isosceles triangle, and thesquare on any line drawn from the vertex to a point in the base, is equal to the rectangle containedby the segments of the base.
6.The square on either side of a right-angled triangle is equal to the rectangle contained by thesum and the difference of the hypotenuse and the other side.
PROP. VI.—Theorem.
PROP. VI.—Theorem.
If a line(AB)be bisected(atC), and divided externally in any point(D), therectangle(AD.BD)contained by the segments made by the external point, togetherwith the square on half the line, is equal to the square on the segment between themiddle point and the point of external division.
Dem.—OnCDdescribe the squareCDFE[I.xlvi.], and joinDE; throughBdrawBHGparallel toCE[I.xxxi.], meetingDEinH; throughHdrawKLMparallel toAD; and throughAdrawAKparallel toCL. Then becauseACis equal toCB, the rectangleALis equal toCH[I.xxxvi.]; but the complementsCH,HFare equal [I.xliii.]; thereforeALis equal toHF. To each of these equals addCMandLG, and we getAMandLGequal to the squareCDFE; butAMis equal to the rectangleAD.DM, and therefore equal to the rectangleAD.DB, sinceDBis equal toDM; alsoLGis equal to the square onCB, andCDFEis the square onCD.Hence therectangleAD.DB, together with the square onCB, is equal to the square onCD.
Or thus:—
Dem.—OnCBdescribe the squareCBEF[I.xlvi.]. JoinBF. ThroughDdrawDGparallel toCF, meetingFBproduced inH. ThroughHdrawKMparallel toAB. ThroughAdrawAKparallel toCL[I.xxxi.].
The parallelogramCMis equal toDE[I.xliii.]; butALis equal toCM[I.xxxvi.], because they are on equal basesAC,CB, and between the same parallels; thereforeALis equal toDE. To each addCH, and we get the parallelogramAHequal to the gnomonCMG; butAHis equal to the rectangleAD.DH, and therefore equal to the rectangleAD.DB, sinceDHis equal toDB[iv.,Cor.1]; therefore the rectangleAD.DBis equal to the gnomonCMG, and the square onCBis the figureCE.Therefore the rectangleAD.DB, together with thesquare onCB, is equal to the whole figureLHGF—that is, equal to the square onLHor to the square onCD.
Exercises.
Exercises.
1.Show that Propositionvi.is reduced to Propositionv.by producing the line in the oppositedirection.
2.Divide a given line externally, so that the rectangle contained by its segments may be equalto the square on a given line.
3.Given the difference of two lines and the rectangle contained by them; find thelines.
4.The rectangle contained by any two lines is equal to the square on half the sum, minus thesquare on half the difference.
5.Given the sum or the difference of two lines and the difference of their squares; find thelines.
6.If from the vertexCof an isosceles triangle a lineCDbe drawn to any point in the baseproduced, prove thatCD2−CB2=AD.DB.
7.Give a common enunciation which will include Propositionsv.andvi.
PROP.VII.—Theorem.
PROP.VII.—Theorem.
If a right line(AB)be divided into any two parts(atC), the sum of the squares onthe whole line(AB)and either segment(CB)is equal to twice the rectangle(2AB.CB)contained by the whole line and that segment, together with the square onthe other segment.
Dem.—OnABdescribe the squareABDE. JoinBE. ThroughCdrawCGparallel toAE, intersectingBEinF. ThroughFdrawHKparallel toAB.
Now the squareADis equal to the three figuresAK,FD, andGH: to each add the squareCK, and we have the sum of the squaresAD,CKequal to the sum of the three figuresAK,CD,GH; butCDis equal toAK; therefore the sum of the squaresAD,CKis equal to twice the figureAK, together with the figureGH. NowAKis the rectangleAB.BK; butBKis equal toBC; thereforeAKis equal to the rectangleAB.BC, andADis the square onAB;CKthe square onCB; andGHis the square onHF, and therefore equal to the square onAC.Hence the sum of the squares onABandBCis equal to twice the rectangleAB.BC, together with the square onAC.
Or thus:OnACdescribe the squareACDE. Produce the sidesCD,DE,EA, and make eachproduced part equal toCB. JoinBF,FG,GH,HB. Then the figureBFGHis a square [I.xlvi.,Ex.3], and it is equal to the square onAC, together with the four equal trianglesHAB,BCF,FDG,GEH. Now [I.xlvii.], the figureBFGHis equal to the sum of the squares onAB,AH—thatis, equal to the sum of the squares onAB,BC; and the sum of the four triangles is equal to twicethe rectangleAB.BC, for each triangle is equal to half the rectangleAB.BC.Hence the sum of thesquares onAB,BCis equal to twice the rectangleAB.BC, together with the square onAC.
Comparison ofiv.andvii.
Byiv., square on sum = sum of squares + twice rectangle.
Byvii., square on difference = sum of squares-twice rectangle.
Cors.fromiv. andvii.
1. Square on the sum, the sum of the squares, and the square on the difference of any two lines, are in arithmetical progression.
2. Square on the sum + square on the difference of any two lines = twice the sum of the squares on the lines (Props.ix. andx.).
3. The square on the sum−the square on the difference of any two lines = four times the rectangle under lines (Prop.viii.).
PROP.VIII.–Theorem.
PROP.VIII.–Theorem.
If a line(AB)be divided into two parts(atC), the square on the sum of the wholeline(AB)and either segment(BC)is equal to four times the rectangle contained bythe whole line(AB)and that segment, together with the square on the other segment(AC).
Dem.—ProduceABtoD. MakeBDequal toBC. OnADdescribe the squareAEFD[I.xlvi.]. JoinDE. ThroughC,BdrawCH,BLparallel toAE[I.xxxi.], and throughK,IdrawMN,POparallel toAD.
SinceCOis the square onCD, andCKthe square onCB, andCBis the half ofCD,COis equal to four timesCK[iv.,Cor.1]. Again, sinceCG,GIare the sides of equal squares, they are equal [I.xlvi.,Cor.1]. Hence the parallelogramAGis equal toMI[I.xxxvi.]. In like mannerILis equal toJF; butMIis equal toIL[I.xliii.]. Therefore the four figuresAG,MI,IL,JFare all equal; hence their sum is equal to four timesAG; and the squareCOhas been proved to be equal to four timesCK. Hence the gnomonAOHis equal to four times the rectangleAK—that is, equal to four times the rectangleAB.BC, sinceBCis equal toBK.
Again, the figurePHis the square onPI, and therefore equal to the square onAC. Hence the whole figureAF, that is,the square onAD, is equal to four times therectangleAB.BC, together with the square onAC.
Or thus:ProduceBAtoD, and makeAD=BC. OnDBdescribe the squareDBEF. Cut offBG,EI,FLeach equal toBC. ThroughAandIdraw lines parallel toDF, and throughGandL,lines parallel toAB.
Now it is evident that the four rectangles.AG,GI,IL,LAare all equal; butAGisthe rectangleAB.BGorAB.BC. Therefore the sum of the four rectangles is equal to4AB.BC. Again, the figureNPis evidently equal to the square onAC. Hence the wholefigure, which is the square onBD, orthe square on the sum ofABandBC, is equal to4AB.BC+AC2.
Direct sequence fromv. orvi.
Since byv. orvi. the rectangle contained by any two lines is = the square on half their sum−the square on half their difference; therefore four times the rectangle contained by any two lines = the square on their sum−the square on their difference.
Direct sequence ofviii. fromiv. andvii.
Byiv., the square on the sum = the sum of the squares + twice the rectangle.
Byvii., the square on the difference = the sum of the squares−twice the rectangle. Therefore, by subtraction, the square on the sum−the square on the difference = four times the rectangle.