 |
| 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.
Proof.—Let a, b, c denote the lines AA′, BB′, CC′, which meet at S. Then these may be taken as bases of projective rows, so that A, A′, S on a correspond to B, B′, S on b, and to C, C′, S on c. As the point S is common to all, any two of these rows will be perspective.
| If | S1 be the centre of projection of rows | b and c, |
| S2 ” ” ” | c and a, | |
| S3 ” ” ” | a and b, |
and if the line S1S2 cuts a in A1, and b in B1, and c in C1, then A1, B1 will be corresponding points in a and b, both corresponding to C1 in c. But a and b are perspective, therefore the line A1B1, that is S1S2, joining corresponding points must pass through the centre of projection S3 of a and b. In other words, S1, S2, S3 lie in a line. This is Desargues’ celebrated theorem if we state it thus:—
Theorem of Desargues.—If each of two triangles has one vertex on each of three concurrent lines, then the intersections of corresponding sides lie in a line, those sides being called corresponding which are opposite to vertices on the same line.
The converse theorem holds also, viz.
Theorem.—If the sides of one triangle meet those of another in three points which lie in a line, then the vertices lie on three lines which meet in a point.
The proof is almost the same as before.
§ 38. Metrical Relations between Projective Rows.—Every row contains one point which is distinguished from all others, viz. the point at infinity. In two projective rows, to the point I at infinity in one corresponds a point I′ in the other, and to the point J′ at infinity in the second corresponds a point J in the first. The points I′ and J are in general finite. If now A and B are any two points in the one, A′, B′ the corresponding points in the other row, then
or
But, by § 17,
therefore the last equation changes into
that is to say—
Theorem.—The product of the distances of any two corresponding points in two projective rows from the points which correspond to the points at infinity in the other is constant, viz. AJ · A′I′ = k. Steiner has called this number k the Power of the correspondence.
[The relation AJ · A′I′ = k shows that if J, I′ be given then the point A′ corresponding to a specified point A is readily found; hence A, A′ generate homographic ranges of which I and J′ correspond to the points at infinity on the ranges. If we take any two origins O, O′, on the ranges and reduce the expression AJ · A′I′ = k to its algebraic equivalent, we derive an equation of the form αxx′ + βx + γx′ + δ = 0. Conversely, if a relation of this nature holds, then points corresponding to solutions in x, x′ form homographic ranges.]
§ 39. Similar Rows.—If the points at infinity in two projective rows correspond so that I′ and J are at infinity, this result loses its meaning. But if A, B, C be any three points in one, A′, B′, C′ the corresponding ones on the other row, we have
which reduces to
that is, corresponding segments are proportional. Conversely, if corresponding segments are proportional, then to the point at infinity in one corresponds the point at infinity in the other. If we call such rows similar, we may state the result thus—
 |
| Fig. 14. |
Theorem.—Two projective rows are similar if to the point at infinity in one corresponds the point at infinity in the other, and conversely, if two rows are similar then they are projective, and the points at infinity are corresponding points.
From this the well-known propositions follow:—
Two lines are cut proportionally (in similar rows) by a series of parallels. The rows are perspective, with centre of projection at infinity.
If two similar rows are placed parallel, then the lines joining homologous points pass through a common point.
§ 40. If two flat pencils be projective, then there exists in either, one single pair of lines at right angles to one another, such that the corresponding lines in the other pencil are again at right angles.
To prove this, we place the pencils in perspective position (fig. 14) by making one ray coincident with its corresponding ray. Corresponding rays meet then on a line p. And now we draw the circle which has its centre O on p, and which passes through the centres S and S′ of the two pencils. This circle cuts p in two points H and K. The two pairs of rays, h, k, and h′, k′, joining these points to S and S′ will be pairs of corresponding rays at right angles. The construction gives in general but one circle, but if the line p is the perpendicular bisector of SS′, there exists an infinite number, and to every right angle in the one pencil corresponds a right angle in the other.
Principle of Duality
§ 41. It has been stated in § 1 that not only points, but also planes and lines, are taken as elements out of which figures are built up. We shall now see that the construction of one figure which possesses certain properties gives rise in many cases to the construction of another figure, by replacing, according to definite rules, elements of one kind by those of another. The new figure thus obtained will then possess properties which may be stated as soon as those of the original figure are known.
We obtain thus a principle, known as the principle of duality or of reciprocity, which enables us to construct to any figure not containing any measurement in its construction a reciprocal figure, as it is called, and to deduce from any theorem a reciprocal theorem, for which no further proof is needed.
It is convenient to print reciprocal propositions on opposite sides of a page broken into two columns, and this plan will occasionally be adopted.
We begin by repeating in this form a few of our former statements:—
| Two points determine a line. | Two planes determine a line. |
| Three points which are not in a line determine a plane. | Three planes which do not pass through a line determine a point. |
| A line and a point without it determine a plane. | A line and a plane not through it determine a point. |
| Two lines in a plane determine a point. | Two lines through a point determine a plane. |
These propositions show that it will be possible, when any figure is given, to construct a second figure by taking planes instead of points, and points instead of planes, but lines where we had lines.
