[p 99 ] II. DEMONSTRATIONS WHICH COULD NOT CONVENIENTLY BE INCLUDED IN THE TEXT . I. THE SECOND COROLLARY, PROBLEM II .

In Fig.8. omit the lines CD, C'D', and DS; and, as here in Fig.75., from a draw ad parallel to AB, cutting BT in d; and from d draw de parallel to BC'.

[Geometric diagram]
Fig.75.

Now as ad is parallel to AB—

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Now because the triangles acV, bc'V, are similar—

But ac is equal to de—

But ad is parallel to AB—

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II.

THE THIRD COROLLARY, PROBLEM III.

In Fig.13., since aR is by construction parallel to AB in Fig.12., and TV is by construction in ProblemIII. also parallel to AB—

Again, by the construction of Fig.13., aR' is parallel to MV—

And it has just been shown that also

But by construction, aR' = aR—

[p102]
III.

ANALYSIS OF PROBLEM XV.

We proceed to take up the general condition of the second problem, before left unexamined, namely, that in which the vertical distances BC' and AC (Fig.6. page 13), as well as the direct distances TD and TD' are unequal.

In Fig.6., here repeated (Fig.76.), produce C'B downwards, and make C'E equal to CA.

[Geometric diagram]
Fig.76.

Join AE.

Then, by the second Corollary of ProblemII., AE is a horizontal line.

Draw TV parallel to AE, cutting the sight-line in V.

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Complete the constructions of ProblemII. and its second Corollary.

Then by ProblemII. ab is the line AB drawn in perspective; and by its Corollary ae is the line AE drawn in perspective.

From V erect perpendicular VP, and produce ab to cut it in P.

Join TP, and from e draw ef parallel to AE, and cutting AT in f.

Now in triangles EBT and AET, as eb is parallel to EB and ef to AE;—eb : ef ? EB : AE[eqnx].

But TV is also parallel to AE and PV to eb.

Therefore also in the triangles aPV and aVT,

Therefore PV : VT ? EB : AE[eqnxii].

And, by construction, angle TPV = ? AEB.

Therefore the triangles TVP, AEB, are similar; and TP is parallel to AB.

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Now the construction in this problem is entirely general for any inclined line AB, and a horizontal line AE in the same vertical plane with it.

So that if we find the vanishing-point of AE in V, and from V erect a vertical VP, and from T draw TP parallel to AB, cutting VP in P, P will be the vanishing-point of AB, and (by the same proof as that given at page 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 from the last figure and repeat it here, Fig.77., adding the measuring-line aM, that the student may observe its position with respect to the other lines before I remove any more of them.

Now if the line AB in this diagram represented the length of the line AB in reality (as AB does in Figs. 10. and 11.), we should only have to proceed to modify Corollary III. of ProblemII. to this new construction. We shall see presently that AB does not represent the actual length of the inclined line AB in nature, nevertheless we shall first proceed as if it did, and modify our result afterwards.

[p105]
In Fig.77. draw ad parallel to AB, cutting BT in d.

Therefore ad is the sight-magnitude of AB, as aR is of AB in Fig.11.

[Geometric diagram]
Fig.78.

Remove again from the figure all lines except PV, VT, PT, ab, ad, and the measuring-line.

Set off on the measuring-line am equal to ad.

Draw PQ parallel to am, and through b draw mQ, cutting PQ in Q.

Then, by the proof already given in page 20, PQ = PT.

Therefore if P is the vanishing-point of an inclined line AB, and QP is a horizontal line drawn through it, make PQ equal to PT, and am on the measuring-line equal to the sight-magnitude of the line AB in the diagram, and the line joining mQ will cut aP in b.

We have now, therefore, to consider what relation the length of the line AB in this diagram, Fig.77., has to the length of the line AB in reality.

Now the line AE in Fig.77. represents the length of AE in reality.

But the angle AEB, 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 from E we draw EC, as in Fig.79., at right angles to AE, make EC = EB, and join AC, AC will be the real length of the line AB.

Now, therefore, if instead of am in Fig.78., we take the real length of AB, that real length will be to am as AC to AB in Fig.79.

And then, if the line drawn to the measuring-line PQ is still to cut aP in b, it is evident that the line PQ must be shortened in the same ratio that am was shortened; and the true dividing-point will be Q' in Fig.80., fixed so that Q'P shall be to QP as am' is to am; am' representing the real length of AB.

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But am' is therefore to am as AC is to AB in Fig.79.

Therefore PQ' must be to PQ as AC is to AB.

But PQ equals PT (Fig.78.); and PV is to VT (in Fig.78.) as BE is to AE (Fig.79.).

Hence we have only to substitute PV for EC, and VT for AE, in Fig.79., and the resulting diagonal AC will be the required length of PQ'.

[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, for VD in Fig.46. (or VM in Fig.15.) =VT by construction in ProblemIV. It should, however, be observed, that the distance PQ' or PX, in Fig.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 in ProblemXX.

THE END.