[p94]PROBLEM XVI.Itis often possible to shorten other perspective operations considerably, by finding the vanishing-points of the inclined lines of the object. Thus, in drawing the gabled roof in Fig. 43., if the gableA Y Cbe drawn in perspective, and the vanishing-point ofA Ydetermined, it is not necessary to draw the two sides of the rectangle,A′ D′andD′ B′, in order to determine the pointY′; but merely to drawY Y′to the vanishing-point ofA A′andA′ Y′to the vanishing-point ofA Y, meeting inY′, the point required.Again, if there be a series of gables, or other figures produced by parallel inclined lines, and retiring to the pointV, as inFig. 72.,[Footnote34]it is not necessary to draw each separately, but merely to determine their breadths on the lineA V, and draw the slopes of each to their vanishing-points, as shown inFig. 72.Or if the gables are equal in height, and a line be drawn fromYtoV, the construction resolves itself into a zigzag drawn alternately toPandQ, between the linesY VandA V.The student must be very cautious, in finding the vanishing-points of inclined lines, to notice their relations to the horizontals beneath them, else he may easily mistake the horizontal to which they belong.Thus, letA B C D,Fig. 73., be a rectangular inclined plane, and let it be required to find the vanishing-point of its diagonalB D.FindV, the vanishing-point ofA DandB C.DrawA Eto the opposite vanishing-point, so thatD A Emay represent a right angle.Let fall fromBthe verticalB E, cuttingA EinE.JoinE D, and produce it to cut the sight-line inV′.[p95][Geometric diagram]Fig. 72.[p96]Then, since the pointEis vertically under the pointB, the horizontal lineE Dis vertically under the inclined lineB D.[Geometric diagram]Fig. 73.So that if we now let fall the verticalV′ PfromV′, and produceB Dto cutV′ PinP, the pointPwill be the vanishing-point ofB D, and of all lines parallel to it.[Footnote35][Footnote34:The diagram is inaccurately cut.Y Vshould be a right line.]Return to text[Footnote35:The student may perhaps understand this construction better by completing the rectangleA D F E, drawingD Fto the vanishing-point ofA E, andE FtoV. The whole figure,B F, may then be conceived as representing half the gable roof of a house,A Fthe rectangle of its base, andA Cthe rectangle of its sloping side.In nearly all picturesque buildings, especially on the Continent, the slopes of gables are much varied (frequently unequal on the two sides), and the vanishing-points of their inclined lines become very important, if accuracy is required in the intersections of tiling, sides of dormer windows, etc.Obviously, also, irregular triangles and polygons in vertical planes may be more easily constructed by finding the vanishing-points of their sides, than by the construction given in thecorollary to Problem IX.; and if such triangles or polygons have others concentrically inscribed within them, as often in Byzantine mosaics, etc., the use of the vanishing-points will become essential.]Return to text
Itis often possible to shorten other perspective operations considerably, by finding the vanishing-points of the inclined lines of the object. Thus, in drawing the gabled roof in Fig. 43., if the gableA Y Cbe drawn in perspective, and the vanishing-point ofA Ydetermined, it is not necessary to draw the two sides of the rectangle,A′ D′andD′ B′, in order to determine the pointY′; but merely to drawY Y′to the vanishing-point ofA A′andA′ Y′to the vanishing-point ofA Y, meeting inY′, the point required.
Again, if there be a series of gables, or other figures produced by parallel inclined lines, and retiring to the pointV, as inFig. 72.,[Footnote34]it is not necessary to draw each separately, but merely to determine their breadths on the lineA V, and draw the slopes of each to their vanishing-points, as shown inFig. 72.Or if the gables are equal in height, and a line be drawn fromYtoV, the construction resolves itself into a zigzag drawn alternately toPandQ, between the linesY VandA V.
The student must be very cautious, in finding the vanishing-points of inclined lines, to notice their relations to the horizontals beneath them, else he may easily mistake the horizontal to which they belong.
Thus, letA B C D,Fig. 73., be a rectangular inclined plane, and let it be required to find the vanishing-point of its diagonalB D.
FindV, the vanishing-point ofA DandB C.
DrawA Eto the opposite vanishing-point, so thatD A Emay represent a right angle.
Let fall fromBthe verticalB E, cuttingA EinE.
JoinE D, and produce it to cut the sight-line inV′.
[p95][Geometric diagram]Fig. 72.
[p96]Then, since the pointEis vertically under the pointB, the horizontal lineE Dis vertically under the inclined lineB D.
[Geometric diagram]Fig. 73.
So that if we now let fall the verticalV′ PfromV′, and produceB Dto cutV′ PinP, the pointPwill be the vanishing-point ofB D, and of all lines parallel to it.[Footnote35]
[Footnote34:The diagram is inaccurately cut.Y Vshould be a right line.]Return to text[Footnote35:The student may perhaps understand this construction better by completing the rectangleA D F E, drawingD Fto the vanishing-point ofA E, andE FtoV. The whole figure,B F, may then be conceived as representing half the gable roof of a house,A Fthe rectangle of its base, andA Cthe rectangle of its sloping side.In nearly all picturesque buildings, especially on the Continent, the slopes of gables are much varied (frequently unequal on the two sides), and the vanishing-points of their inclined lines become very important, if accuracy is required in the intersections of tiling, sides of dormer windows, etc.Obviously, also, irregular triangles and polygons in vertical planes may be more easily constructed by finding the vanishing-points of their sides, than by the construction given in thecorollary to Problem IX.; and if such triangles or polygons have others concentrically inscribed within them, as often in Byzantine mosaics, etc., the use of the vanishing-points will become essential.]Return to text
[Footnote34:The diagram is inaccurately cut.Y Vshould be a right line.]Return to text
[Footnote35:The student may perhaps understand this construction better by completing the rectangleA D F E, drawingD Fto the vanishing-point ofA E, andE FtoV. The whole figure,B F, may then be conceived as representing half the gable roof of a house,A Fthe rectangle of its base, andA Cthe rectangle of its sloping side.
In nearly all picturesque buildings, especially on the Continent, the slopes of gables are much varied (frequently unequal on the two sides), and the vanishing-points of their inclined lines become very important, if accuracy is required in the intersections of tiling, sides of dormer windows, etc.
Obviously, also, irregular triangles and polygons in vertical planes may be more easily constructed by finding the vanishing-points of their sides, than by the construction given in thecorollary to Problem IX.; and if such triangles or polygons have others concentrically inscribed within them, as often in Byzantine mosaics, etc., the use of the vanishing-points will become essential.]Return to text
[p97]PROBLEM XVIII.Beforeexamining the last three problems it is necessary that you should understand accurately what is meant by the position of an inclined plane.Cut a piece of strong white pasteboard into any irregular shape, and dip it in a sloped position into water. However you hold it, the edge of the water, of course, will always draw a horizontal line across its surface. The direction of this horizontal line is the direction of the inclined plane. (In beds of rock geologists call it their “strike.”)[Geometric diagram]Fig. 74.Next, draw a semicircle on the piece of pasteboard; draw its diameter,A B,Fig. 74., and a vertical line from its center,C D; and draw some other lines,C E,C F, etc., from the center to any points in the circumference.Now dip the piece of pasteboard again into water, and, holding it at any inclination and in any direction you choose, bring the surface of the water to the lineA B. Then the lineC Dwill be the most steeply inclined of all the lines drawn to the circumference of the circle;G CandH Cwill be less steep; andE CandF Cless steep still. The nearer the lines toC D, the steeper they will be; and the nearer toA B, the more nearly horizontal.[p98]When, therefore, the lineA Bis horizontal (or marks the water surface), its direction is the direction of the inclined plane, and the inclination of the lineD Cis the inclination of the inclined plane. In beds of rock geologists call the inclination of the lineD Ctheir “dip.”To fix the position of an inclined plane, therefore, is to determine the direction of any two lines in the plane,A BandC D, of which one shall be horizontal and the other at right angles to it. Then any lines drawn in the inclined plane, parallel toA B, will be horizontal; and lines drawn parallel toC Dwill be as steep asC D, and are spoken of in the text as the “steepest lines” in the plane.But farther, whatever the direction of a plane may be, if it be extended indefinitely, it will be terminated, to the eye of the observer, by a boundary line, which, in a horizontal plane, is horizontal (coinciding nearly with the visible horizon);—in a vertical plane, is vertical;—and, in an inclined plane, is inclined.This line is properly, in each case, called the “sight-line” of such plane; but it is only properly called the “horizon” in the case of a horizontal plane: and I have preferred using always the term “sight-line,” not only because more comprehensive, but more accurate; for though the curvature of the earth’s surface is so slight that practically its visible limit always coincides with the sight-line of a horizontal plane, it does not mathematically coincide with it, and the two lines ought not to be considered as theoretically identical, though they are so in practice.It is evident that all vanishing-points of lines in any plane must be found on its sight-line, and, therefore, that the sight-line of any plane may be found by joining any two of such vanishing-points. Hence the construction ofProblem XVIII.
Beforeexamining the last three problems it is necessary that you should understand accurately what is meant by the position of an inclined plane.
Cut a piece of strong white pasteboard into any irregular shape, and dip it in a sloped position into water. However you hold it, the edge of the water, of course, will always draw a horizontal line across its surface. The direction of this horizontal line is the direction of the inclined plane. (In beds of rock geologists call it their “strike.”)
[Geometric diagram]Fig. 74.
Next, draw a semicircle on the piece of pasteboard; draw its diameter,A B,Fig. 74., and a vertical line from its center,C D; and draw some other lines,C E,C F, etc., from the center to any points in the circumference.
Now dip the piece of pasteboard again into water, and, holding it at any inclination and in any direction you choose, bring the surface of the water to the lineA B. Then the lineC Dwill be the most steeply inclined of all the lines drawn to the circumference of the circle;G CandH Cwill be less steep; andE CandF Cless steep still. The nearer the lines toC D, the steeper they will be; and the nearer toA B, the more nearly horizontal.
[p98]When, therefore, the lineA Bis horizontal (or marks the water surface), its direction is the direction of the inclined plane, and the inclination of the lineD Cis the inclination of the inclined plane. In beds of rock geologists call the inclination of the lineD Ctheir “dip.”
To fix the position of an inclined plane, therefore, is to determine the direction of any two lines in the plane,A BandC D, of which one shall be horizontal and the other at right angles to it. Then any lines drawn in the inclined plane, parallel toA B, will be horizontal; and lines drawn parallel toC Dwill be as steep asC D, and are spoken of in the text as the “steepest lines” in the plane.
But farther, whatever the direction of a plane may be, if it be extended indefinitely, it will be terminated, to the eye of the observer, by a boundary line, which, in a horizontal plane, is horizontal (coinciding nearly with the visible horizon);—in a vertical plane, is vertical;—and, in an inclined plane, is inclined.
This line is properly, in each case, called the “sight-line” of such plane; but it is only properly called the “horizon” in the case of a horizontal plane: and I have preferred using always the term “sight-line,” not only because more comprehensive, but more accurate; for though the curvature of the earth’s surface is so slight that practically its visible limit always coincides with the sight-line of a horizontal plane, it does not mathematically coincide with it, and the two lines ought not to be considered as theoretically identical, though they are so in practice.
It is evident that all vanishing-points of lines in any plane must be found on its sight-line, and, therefore, that the sight-line of any plane may be found by joining any two of such vanishing-points. Hence the construction ofProblem XVIII.
[p99]II.DEMONSTRATIONS WHICH COULD NOT CONVENIENTLY BE INCLUDEDIN THE TEXT.I.THE SECOND COROLLARY,PROBLEM II.InFig. 8.omit the linesC D,C′ D′, andD S; and, as here inFig. 75., fromadrawa dparallel toA B, cuttingB Tind; and fromddrawd eparallel toB C′.[Geometric diagram]Fig. 75.Now asa dis parallel toA B—A C ∶a c∷ B C′∶d e;[eqniii]butA Cis equal toB C′—∴a c=d e.[p100]Now because the trianglesa cV,b c′V, aresimilar—a c∶b c′∷aV ∶bV;[eqniv]and because the trianglesd eT,b c′Taresimilar—d e∶b c′∷dT ∶bT.[eqnv]Buta cis equal tod e—∴aV ∶bV ∷dT ∶bT[eqnvi];∴the two trianglesa b d,bT V, are similar, and their angles are alternate;∴ T Vis parallel toa d.Buta dis parallel toA B—∴ T Vis parallel toA B.
InFig. 8.omit the linesC D,C′ D′, andD S; and, as here inFig. 75., fromadrawa dparallel toA B, cuttingB Tind; and fromddrawd eparallel toB C′.
[Geometric diagram]Fig. 75.
Now asa dis parallel toA B—
A C ∶a c∷ B C′∶d e;[eqniii]butA Cis equal toB C′—∴a c=d e.
butA Cis equal toB C′—
[p100]Now because the trianglesa cV,b c′V, aresimilar—
a c∶b c′∷aV ∶bV;[eqniv]and because the trianglesd eT,b c′Taresimilar—d e∶b c′∷dT ∶bT.[eqnv]
and because the trianglesd eT,b c′Taresimilar—
Buta cis equal tod e—
∴aV ∶bV ∷dT ∶bT[eqnvi];∴the two trianglesa b d,bT V, are similar, and their angles are alternate;∴ T Vis parallel toa d.
∴the two trianglesa b d,bT V, are similar, and their angles are alternate;
Buta dis parallel toA B—
∴ T Vis parallel toA B.
[p101]II.THE THIRD COROLLARY,PROBLEM III.InFig. 13., sinceaRis by construction parallel toA BinFig. 12., andT Vis by construction inProblem III.also parallel toA B—∴aRis parallel toT V,∴a bRandTbVare alternate triangles,∴aR ∶ T V ∷a b∶bV[eqnvii].Again, by the construction ofFig. 13.,aR′is parallel toM V—∴a bR′andMbVare alternate triangles,∴aR′ ∶ M V ∷a b∶bV[eqnviii].And it has just been shown that alsoaR ∶ T V ∷a b∶bV—∴aR′ ∶ M V ∷aR ∶ T V[eqnix].But by construction,aR′=aR—∴ M V = T V.
InFig. 13., sinceaRis by construction parallel toA BinFig. 12., andT Vis by construction inProblem III.also parallel toA B—
∴aRis parallel toT V,∴a bRandTbVare alternate triangles,∴aR ∶ T V ∷a b∶bV[eqnvii].
Again, by the construction ofFig. 13.,aR′is parallel toM V—
∴a bR′andMbVare alternate triangles,∴aR′ ∶ M V ∷a b∶bV[eqnviii].
And it has just been shown that also
aR ∶ T V ∷a b∶bV—∴aR′ ∶ M V ∷aR ∶ T V[eqnix].
But by construction,aR′=aR—
∴ M V = T V.
[p102]III.ANALYSIS OFPROBLEM XV.Weproceed to take up the general condition of the second problem, before left unexamined, namely, that in which the vertical distancesB C′andA C(Fig. 6.page 13), as well as the direct distancesT DandT D′are unequal.InFig. 6., here repeated (Fig. 76.), produceC′ Bdownwards, and makeC′ Eequal toC A.[Geometric diagram]Fig. 76.JoinA E.Then, by the secondCorollaryofProblem II.,A Eis a horizontal line.DrawT Vparallel toA E, cutting the sight-line inV.∴ Vis the vanishing-point ofA E.[p103]Complete the constructions ofProblem II.and its secondCorollary.Then byProblem II.a bis the lineA Bdrawn in perspective; and by itsCorollarya eis the lineA Edrawn in perspective.FromVerect perpendicularV P, and producea bto cut it inP.JoinT P, and fromedrawe fparallel toA E, and cuttingA Tinf.Now in trianglesE B TandA E T, ase bis parallel toE Bande ftoA E;—e b∶e f∷ E B ∶ A E[eqnx].ButT Vis also parallel toA EandP Vtoe b.Therefore also in the trianglesaP VandaV T,e b∶e f∷ P V ∶ V T[eqnxi].ThereforeP V ∶ V T ∷ E B ∶ A E[eqnxii].And, by construction, angleT P V = ∠ A E B.Therefore the trianglesT V P,A E B, are similar; andT Pis parallel toA B.[p104]Now the construction in this problem is entirely general for any inclined lineA B, and a horizontal lineA Ein the same vertical plane with it.So that if we find the vanishing-point ofA EinV, and fromVerect a verticalV P, and fromTdrawT Pparallel toA B, cuttingV PinP,Pwill be the vanishing-point ofA B, and (by the same proof as that given atpage 17) of all lines parallel to it.[Geometric diagram]Fig. 77.Next, to find the dividing-point of the inclined line.I remove some unnecessary lines fromthe last figureand repeat it here,Fig. 77., adding the measuring-lineaM, that the student may observe its position with respect to the other lines before I remove any more of them.Now if the lineA Bin this diagram represented the length of the lineA Bin reality (asA BdoesinFigs. 10.and11.), we should only have to proceed to modifyCorollary III.ofProblem II.to this new construction. We shall see presently thatA Bdoes not represent the actual length of the inclined lineA Bin nature, nevertheless we shall first proceed as if it did, and modify our result afterwards.[p105]InFig. 77.drawa dparallel toA B, cuttingB Tind.Thereforea dis the sight-magnitude ofA B, asaRis ofA BinFig. 11.[Geometric diagram]Fig. 78.Remove again from the figure all lines exceptP V,V T,P T,a b,a d, and the measuring-line.Set off on the measuring-linea mequal toa d.DrawP Qparallel toa m, and throughbdrawmQ, cuttingP QinQ.Then, by the proof already given inpage 20,P Q=P T.Therefore ifPis the vanishing-point of an inclined lineA B, andQ Pis a horizontal line drawn through it, makeP Qequal toP T, anda mon the measuring-line equal to the sight-magnitude of the lineA Bin the diagram, and the line joiningmQwill cutaPinb.We have now, therefore, to consider what relation the length of the lineA Bin this diagram,Fig. 77., has to the length of the lineA Bin reality.Now the lineA EinFig. 77.represents the length ofA Ein reality.But the angleA E B,Fig. 77., and the corresponding angle in all the constructions of the earlier problems, is in reality a right angle, though in the diagram necessarily represented as obtuse.[Geometric diagram]Fig. 79.Therefore, if fromEwe drawE C, as inFig. 79., at right angles toA E, makeE C=E B, and joinA C,A Cwill be the real length of the lineA B.Now, therefore, if instead ofa minFig. 78., we take the real length ofA B, that real length will be toa masA CtoA BinFig. 79.And then, if the line drawn to the measuring-lineP Qis still to cutaPinb, it is evident that the lineP Qmust be shortened in the same ratio thata mwas shortened; and the true dividing-point will beQ′inFig. 80., fixed so thatQ′ Pshall be toQ Pasa m′is toa m;a m′representing the real length ofA B.[p106]Buta m′is therefore toa masA Cis toA BinFig. 79.ThereforeP Q′must be toP QasA Cis toA B.ButP QequalsP T(Fig. 78.); andP Vis toV T(inFig. 78.) asB Eis toA E(Fig. 79.).Hence we have only to substituteP VforE C, andV TforA E, inFig. 79., and the resulting diagonalA Cwill be the required length ofP Q′.[Geometric diagram]Fig. 80.It will be seen that the construction given in the text (Fig. 46.) is the simplest means of obtaining this magnitude, forV DinFig. 46.(orV MinFig. 15.) =V Tby construction inProblem IV. It should, however, be observed, that the distanceP Q′orP X, inFig. 46., may be laid on the sight-line of the inclined plane itself, if the measuring-line be drawn parallel to that sight-line. And thus any form may be drawn on an inclined plane as conveniently as on a horizontal one, with the single exception of the radiation of the verticals, which have a vanishing-point, as shown inProblem XX.THE END.
Weproceed to take up the general condition of the second problem, before left unexamined, namely, that in which the vertical distancesB C′andA C(Fig. 6.page 13), as well as the direct distancesT DandT D′are unequal.
InFig. 6., here repeated (Fig. 76.), produceC′ Bdownwards, and makeC′ Eequal toC A.
[Geometric diagram]Fig. 76.
JoinA E.
Then, by the secondCorollaryofProblem II.,A Eis a horizontal line.
DrawT Vparallel toA E, cutting the sight-line inV.
∴ Vis the vanishing-point ofA E.
[p103]Complete the constructions ofProblem II.and its secondCorollary.
Then byProblem II.a bis the lineA Bdrawn in perspective; and by itsCorollarya eis the lineA Edrawn in perspective.
FromVerect perpendicularV P, and producea bto cut it inP.
JoinT P, and fromedrawe fparallel toA E, and cuttingA Tinf.
Now in trianglesE B TandA E T, ase bis parallel toE Bande ftoA E;—e b∶e f∷ E B ∶ A E[eqnx].
ButT Vis also parallel toA EandP Vtoe b.
Therefore also in the trianglesaP VandaV T,
e b∶e f∷ P V ∶ V T[eqnxi].
ThereforeP V ∶ V T ∷ E B ∶ A E[eqnxii].
And, by construction, angleT P V = ∠ A E B.
Therefore the trianglesT V P,A E B, are similar; andT Pis parallel toA B.
[p104]Now the construction in this problem is entirely general for any inclined lineA B, and a horizontal lineA Ein the same vertical plane with it.
So that if we find the vanishing-point ofA EinV, and fromVerect a verticalV P, and fromTdrawT Pparallel toA B, cuttingV PinP,Pwill be the vanishing-point ofA B, and (by the same proof as that given atpage 17) of all lines parallel to it.
[Geometric diagram]Fig. 77.
Next, to find the dividing-point of the inclined line.
I remove some unnecessary lines fromthe last figureand repeat it here,Fig. 77., adding the measuring-lineaM, that the student may observe its position with respect to the other lines before I remove any more of them.
Now if the lineA Bin this diagram represented the length of the lineA Bin reality (asA BdoesinFigs. 10.and11.), we should only have to proceed to modifyCorollary III.ofProblem II.to this new construction. We shall see presently thatA Bdoes not represent the actual length of the inclined lineA Bin nature, nevertheless we shall first proceed as if it did, and modify our result afterwards.
[p105]InFig. 77.drawa dparallel toA B, cuttingB Tind.
Thereforea dis the sight-magnitude ofA B, asaRis ofA BinFig. 11.
[Geometric diagram]Fig. 78.
Remove again from the figure all lines exceptP V,V T,P T,a b,a d, and the measuring-line.
Set off on the measuring-linea mequal toa d.
DrawP Qparallel toa m, and throughbdrawmQ, cuttingP QinQ.
Then, by the proof already given inpage 20,P Q=P T.
Therefore ifPis the vanishing-point of an inclined lineA B, andQ Pis a horizontal line drawn through it, makeP Qequal toP T, anda mon the measuring-line equal to the sight-magnitude of the lineA Bin the diagram, and the line joiningmQwill cutaPinb.
We have now, therefore, to consider what relation the length of the lineA Bin this diagram,Fig. 77., has to the length of the lineA Bin reality.
Now the lineA EinFig. 77.represents the length ofA Ein reality.
But the angleA E B,Fig. 77., and the corresponding angle in all the constructions of the earlier problems, is in reality a right angle, though in the diagram necessarily represented as obtuse.
[Geometric diagram]Fig. 79.
Therefore, if fromEwe drawE C, as inFig. 79., at right angles toA E, makeE C=E B, and joinA C,A Cwill be the real length of the lineA B.
Now, therefore, if instead ofa minFig. 78., we take the real length ofA B, that real length will be toa masA CtoA BinFig. 79.
And then, if the line drawn to the measuring-lineP Qis still to cutaPinb, it is evident that the lineP Qmust be shortened in the same ratio thata mwas shortened; and the true dividing-point will beQ′inFig. 80., fixed so thatQ′ Pshall be toQ Pasa m′is toa m;a m′representing the real length ofA B.
[p106]Buta m′is therefore toa masA Cis toA BinFig. 79.
ThereforeP Q′must be toP QasA Cis toA B.
ButP QequalsP T(Fig. 78.); andP Vis toV T(inFig. 78.) asB Eis toA E(Fig. 79.).
Hence we have only to substituteP VforE C, andV TforA E, inFig. 79., and the resulting diagonalA Cwill be the required length ofP Q′.
[Geometric diagram]Fig. 80.
It will be seen that the construction given in the text (Fig. 46.) is the simplest means of obtaining this magnitude, forV DinFig. 46.(orV MinFig. 15.) =V Tby construction inProblem IV. It should, however, be observed, that the distanceP Q′orP X, inFig. 46., may be laid on the sight-line of the inclined plane itself, if the measuring-line be drawn parallel to that sight-line. And thus any form may be drawn on an inclined plane as conveniently as on a horizontal one, with the single exception of the radiation of the verticals, which have a vanishing-point, as shown inProblem XX.
THE END.
Transcriber’s NoteA handful of unequivocal typographical errors has been corrected.For increased clarity, a few diagrams have been shifted from their original position in the text.Images for sections of the text where the∶ratio and∷proportion symbols occur.[i]P′ Q′ ∶ P Q ∷ S T ∶ D TReturn to text[ii]A T ∶ a T ∷ B T ∶ b TReturn to text[iii]A C ∶ a c ∷ B C′ ∶ d eReturn to text[iv]a c ∶ b c′ ∷ a V ∶ b VReturn to text[v]d e ∶ b c′ ∷ d T ∶ b TReturn to text[vi]∴ a V ∶ b V ∷ d T ∶ b TReturn to text[vii]∴ a R ∶ T V ∷ a b ∶ b VReturn to text[viii]∴ a R′ ∶ M V ∷ a b ∶ b VReturn to text[ix]a R ∶ T V ∷ a b ∶ b V— ∴ a R′ ∶ M V ∷ a R ∶ T VReturn to text[x]e b ∶ e f ∷ E B ∶ A EReturn to text[xi]e b ∶ e f ∷ P V ∶ V TReturn to text[xii]P V ∶ V T ∷ E B ∶ A EReturn to text
A handful of unequivocal typographical errors has been corrected.
For increased clarity, a few diagrams have been shifted from their original position in the text.
Images for sections of the text where the∶ratio and∷proportion symbols occur.[i]P′ Q′ ∶ P Q ∷ S T ∶ D TReturn to text[ii]A T ∶ a T ∷ B T ∶ b TReturn to text[iii]A C ∶ a c ∷ B C′ ∶ d eReturn to text[iv]a c ∶ b c′ ∷ a V ∶ b VReturn to text[v]d e ∶ b c′ ∷ d T ∶ b TReturn to text[vi]∴ a V ∶ b V ∷ d T ∶ b TReturn to text[vii]∴ a R ∶ T V ∷ a b ∶ b VReturn to text[viii]∴ a R′ ∶ M V ∷ a b ∶ b VReturn to text[ix]a R ∶ T V ∷ a b ∶ b V— ∴ a R′ ∶ M V ∷ a R ∶ T VReturn to text[x]e b ∶ e f ∷ E B ∶ A EReturn to text[xi]e b ∶ e f ∷ P V ∶ V TReturn to text[xii]P V ∶ V T ∷ E B ∶ A EReturn to text
Images for sections of the text where the∶ratio and∷proportion symbols occur.