S; or, in other words, that the punctuated lines PT' and PR' are in perspective. Thus, if the transversal o,d, be drawn parallel to PT, we have

and if the transversal 0,8 be drawn parallel to PR', we have

Whereupon, multiplying together the right and left hand members of these three equations, we obtain

Again, if o,'l be drawn parallel to AC or PR, we have

(6).

=n, a constant; then from equation (7)

and the lines e,e,' and d1d,' meet in the same point on line 0,0, which is therefore the centre of perspective S.

Q. E. D.

The perspectivity of the punctuated lines PT' and PR' may be deduced from the projective relations of Fig. 15 by the aid of the following two Lemmas, already proved in Art. 1, ch. i.

I.—If the correlative or homologous points a and ɑ, of the punctuated lines u and u, (Fig. 16) coincide with their point of inter- er

homologous points, and let bb1, cc, meet in S; then S will be the centre of perspective of the two lines, seeing that in a harmonic or other system of ratios three members of a compound proportion consisting of four terms are sufficient to determine the fourth. Then the fourth point and any other two of the known points will serve to determine a fifth; and so on ad indefinitum (§ 3).

и

d

di

A

Fig.

a

17.

102

II. Similarly, if in Fig. 17 any pair of homologous rays Sa and S1a, of two different pencils in the same plane Bu are coincident or COperspective, the pencils will be perspective of the same line and consequently in perspective. For, let the homologous rays Sb and S1b, meet in a point B, and the rays

Sc and S,c, in C; then the line BC will be perspective of each pencil; inasmuch as, if three homologous rays of the pencils meet upon the line BC, it necessarily follows that a fourth pair of homologous rays will meet upon the same line.

Now it will be observed that lines PT and PR (Fig. 15) are in perspective, being projected from infinity on the right of the figure; so also are lines PT and PT' as projected from pole B, as well as lines PR and PR' as projected from pole C. Further, the point P is a double point common to lines PT' and PR' as projected from poles B and C respectively; for P does not change its position when projected through pole B from line PT to line PT'; nor does it change when projected, first from infinity parallel to dd', ee', etc. from line PT to line PR, and thence through pole C upon line PR'. Moreover, the point P is the point of intersection of lines PT' and PR'; therefore, by Lemma I., the lines PT' and PR' are in perspective.

In order to find the centre of perspective of lines PT' and PR', we have the point k on line PT correlative of the point at infinity on PT', k being determined by drawing Bk parallel to PT. The point k,' on PR' correlative of k on PT is found as before by projecting k first upon PR in k' and thence upon PR in k'. Thus, the required centre of perspective must be somewhere on the indefinite line joining k' with the point at infinity upon PT. Again, the point on PR corresponds to the point at infinity on PR', to point i on PT, and to point i, on PT'; hence the sought centre must also lie somewhere on the indefinite line joining i, to the point at infinity on PR'; wherefore it coincides with the intersection of lines i, S and k,'S.

Similarly, it can be shewn that the lines PT and PR' are in perspective; for the line drawn from C, the pole of projection for line PR', to S∞, the pole of projection for line PT; or, in other words, the line drawn from C parallel to dd', ee', etc. is a homologous ray common to both lines; therefore, according to Lemma II., the lines PT and PR' are in perspective. The corresponding centre of perspective is determined as follows. When the line oo',

moving parallel to itself, passes to infinity; or, in other terms, when the points o and o' pass to infinity on lines PT and PR, the ray Co' takes up the position CF, parallel to PR. Hence F is the point on line PR' corresponding to infinity on PT; wherefore the required centre must lie somewhere on the line through F parallel to PT. But i is the point on PT corresponding to infinity on PR', therefore the centre must also lie somewhere on the line through i parallel to PR'; so that it lies at the intersection of lines FG and iG.

SOLUTION II.-Newton has given a second solution of the same problem. Thus, let A, B, C, P, Q (Fig. 18) be the A,B,C, given five points. Join any three A, B, C. Taking B and C as fixed poles, draw the line Bp so that angle PBp may be equal to ABa; and Cp so that angle PCp may be equal to ACa. Let Bp and Cp intersect in p. Again, draw the line Bq, making angle QBq equal to ABa; and Cq, making angle QCq equal to ACa. Let Bq and Cq intersect in q Then, if the limb Ba of the fork ABa and the limb Ca of the fork ACa concurrently describe the line pq in the manner just explained, the second limbs of the same forks,

namely BA and CA, will by their intersection describe the conic.

PROOF.-By construction we have the angles PBp and QBq each equal to ABa, and the angles PCP and QCq each equal to ACa. Now from P draw the lines PT and PS, making

angle BPT=Bpq and angle CPS = Cpq,

and let these two lines intersect BQ and CQ in points T and S respectively. Then, if from the equal fork angles PBp and QBq we take away the common angle PBq, the remaining angles pBq and PBQ will be equal.

Similarly, if from the equal fork angles PCp and QCq we subtract the common angle PCq, the remaining angles pCq and PCS will be equal.

Thus, in the triangles pBq and PBT, the angle pBq is equal to PBQ or PBT and by construction angle BPT is equal to Bpq; wherefore the remaining two angles are equal and the triangles similar. Hence

Now the points BCPp are fixed, so that the ratios

are constant; whence it follows that their equals

PC

PC

PT

PS

PS

and are constant, as likewise the ratio

Conse

pq

pq

PT

quently, if R be a sixth point on the conic, we must have

so that the lines PT and PS are similarly divided by the lower prongs of the forks revolving about B and C.

COR. I. Join BP (Fig. 19) and set off Bp so that