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.

[p104]
]
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.