II.

II.

DEMONSTRATIONS WHICH COULD NOT CONVENIENTLY BE INCLUDED IN THE TEXT.

I.

THE SECOND COROLLARY, PROBLEM II.

In Fig. 8. omit the lines C D, C′ D′, and D S; and, as here in Fig. 75., from a draw a d parallel to A B, cutting B T in d; and from d draw d e parallel to B C′.

Fig. 75.

Now as a d is parallel to A B—

A C ∶

a c

∷ B C′

d e

;

[

eqn

iii]

but A C is equal to B C′—

∴

a c

=

d e

.

Now because the triangles a c V, b c′ V, are similar—

a c

∶

b c′

∷

a

V ∶

b

V

;

[

eqn

iv]

and because the triangles d e T, b c′ T are similar—

d e

∶

b c′

∷

d

T ∶

b

T

.

[

eqn

v]

But a c is equal to d e—

∴

a

V ∶

b

V ∷

d

T ∶

b

T

[

eqn

vi]

;

∴ the two triangles a b d, b T V, are similar, and their angles are alternate;

∴ T V

is parallel to

a d

.

But a d is parallel to A B—

∴ T V is parallel to A B.

II.

THE THIRD COROLLARY, PROBLEM III.

In Fig. 13., since a R is by construction parallel to A B in Fig. 12., and T V is by construction in Problem III. also parallel to A B—

∴ a R is parallel to T V, ∴ a b R and T b V are alternate triangles, ∴ a R ∶ T V ∷ a b ∶ b V[eqn vii].

Again, by the construction of Fig. 13., a R′ is parallel to M V—

∴ a b R′ and M b V are alternate triangles, ∴ a R′ ∶ M V ∷ a b ∶ b V[eqn viii].

And it has just been shown that also

a R ∶ T V ∷ a b ∶ b V— ∴ a R′ ∶ M V ∷ a R ∶ T V[eqn ix].

But by construction, a R′ = a R—

∴ M V = T V.

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 B C′ and A C (Fig. 6. page 13), as well as the direct distances T D and T D′ are unequal.

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

Fig. 76.

Join A E.

Then, by the second Corollary of Problem II., A E is a horizontal line.

Draw T V parallel to A E, cutting the sight-line in V.

∴ V is the vanishing-point of A E.

Complete the constructions of Problem II. and its second Corollary.

Then by Problem II. a b is the line A B drawn in perspective; and by its Corollary a e is the line A E drawn in perspective.

From V erect perpendicular V P, and produce a b to cut it in P.

Join T P, and from e draw e f parallel to A E, and cutting A T in f.

Now in triangles E B T and A E T, as e b is parallel to E B and e f to A E;—e b ∶ e f ∷ E B ∶ A E[eqn x].

But T V is also parallel to A E and P V to e b.

Therefore also in the triangles a P V and a V T,

e b ∶ e f ∷ P V ∶ V T[eqn xi].

Therefore P V ∶ V T ∷ E B ∶ A E[eqn xii].

And, by construction, angle T P V = ∠ A E B.

Therefore the triangles T V P, A E B, are similar; and T P is parallel to A B.

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

So that if we find the vanishing-point of A E in V, and from V erect a vertical V P, and from T draw T P parallel to A B, cutting V P in P, P will be the vanishing-point of A B, and (by the same proof as that given at page 17) of all lines parallel to it.

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 a M, that the student may observe its position with respect to the other lines before I remove any more of them.

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

In Fig. 77. draw a d parallel to A B, cutting B T in d.

Therefore a d is the sight-magnitude of A B, as a R is of A B in Fig. 11.

Fig. 78.

Remove again from the figure all lines except P V, V T, P T, a b, a d, and the measuring-line.

Set off on the measuring-line a m equal to a d.

Draw P Q parallel to a m, and through b draw m Q, cutting P Q in Q.

Then, by the proof already given in page 20, P Q = P T.

Therefore if P is the vanishing-point of an inclined line A B, and Q P is a horizontal line drawn through it, make P Q equal to P T, and a m on the measuring-line equal to the sight-magnitude of the line A B in the diagram, and the line joining m Q will cut a P in b.

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

Now the line A E in Fig. 77. represents the length of A E in reality.

But the angle A 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.

Fig. 79.

Therefore, if from E we draw E C, as in Fig. 79., at right angles to A E, make E C = E B, and join A C, A C will be the real length of the line A B.

Now, therefore, if instead of a m in Fig. 78., we take the real length of A B, that real length will be to a m as A C to A B in Fig. 79.

And then, if the line drawn to the measuring-line P Q is still to cut a P in b, it is evident that the line P Q must be shortened in the same ratio that a m was shortened; and the true dividing-point will be Q′ in Fig. 80., fixed so that Q′ PQ P as a m′ is to a m; a m′ representing the real length of A B.

But a m′ is therefore to a m as A C is to A B in Fig. 79.

Therefore P Q′ must be to P Q as A C is to A B.

But P Q equals P T (Fig. 78.); and P V is to V T (in Fig. 78.) as B E is to A E (Fig. 79.).

Hence we have only to substitute P V for E C, and V T for A E, in Fig. 79., and the resulting diagonal A C will be the required length of P Q′.

Fig. 80.

It will be seen that the construction given in the text (Fig. 46.) is the simplest means of obtaining this magnitude, for V D in Fig. 46. (or V M in Fig. 15.) = V T by construction in Problem IV. It should, however, be observed, that the distance P Q′ or P X, 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 Problem XX.

THE END.