The solution is made to depend on the construction of an auxiliary row or pencil which is perspective to both the given ones. This is found as follows:—
Solution of Problem I.—On the line joining two corresponding points, say AA′ (fig. 11), take any two points, S and S′, as centres of auxiliary pencils. Join the intersection B1 of SB and S′B′ to the intersection C1 of SC and S′C′ by the line s1. Then a row on s1 will be perspective to s with S as centre of projection, and to s′ with S′ as centre. To find now the point D′ on s′ corresponding to a point D on s we have only to determine the point D1, where the line SD cuts s1, and to draw S′D1; the point where this line cuts s′ will be the required point D′.
Proof.—The rows s and s′ are both perspective to the row s1, hence they are projective to one another. To A, B, C, D on s correspond A1, B1, C1, D1 on s1, and to these correspond A′, B′, C′, D′ on s′; so that D and D′ are corresponding points as required.
| Fig. 12. |
| Fig. 13. |
Solution of Problem II.—Through the intersection A of two corresponding rays a and a′ (fig. 12), take two lines, s and s′, as bases of auxiliary rows. Let S1 be the point where the line b1, which joins B and B′, cuts the line c1, which joins C and C′. Then a pencil S1 will be perspective to S with s as axis of projection. To find the ray d′ in S′ corresponding to a given ray d in S, cut d by s at D; project this point from S1 to D′ on s′ and join D′ to S′. This will be the required ray.
Proof.—That the pencil S1 is perspective to S and also to S′ follows from construction. To the lines a1, b1, c1, d1 in S1 correspond the lines a, b, c, d in S and the lines a′, b′, c′, d′ in S′, so that d and d′ are corresponding rays.
In the first solution the two centres, S, S′, are any two points on a line joining any two corresponding points, so that the solution of the problem allows of a great many different constructions. But whatever construction be used, the point D′, corresponding to D, must be always the same, according to the theorem in § 29. This gives rise to a number of theorems, into which, however, we shall not enter. The same remarks hold for the second problem.
§ 37. Homological Triangles.—As a further application of the theorems about perspective rows and pencils we shall prove the following important theorem.
Theorem.—If ABC and A′B′C′ (fig. 13) be two triangles, such that the lines AA′, BB′, CC′ meet in a point S, then the intersections of BC and B′C′, of CA and C′A′, and of AB and A′B′ will lie in a line. Such triangles are said to be homological, or in perspective. The triangles are “co-axial” in virtue of the property that the meets of corresponding sides are collinear and copolar, since the lines joining corresponding vertices are concurrent.