straight line rays, such as GFF' and GCC', KFF" and KCC", HF'F" and HC'C", be drawn, so as to determine by their mutual intersections the corresponding triangles FFF" and CC'C"; then the corresponding lines FC, F'C', and F"C" produced will meet in the same point I. For, let FF' and CC' meet in G; then, since the point I of intersection of the lines uu and u'u' is a homologous point common to those lines, the lines uu and u'u' are in perspective from the centre G (Lemma I.). For similar reasons the lines u'u' and u"u" are in perspective from the centre H, the point of intersection of the sides. F'F" and C'C". But, the three pairs of points FF", CC", and PP" being in perspective of the three points F'C'P' respectively, the fourth pair, or double point of intersection II", must be in perspective of the point of intersection I'; or, in other words, the three points of intersection coincide.

Again, if the points II'I" coincide and the pairs FF' and CC', FF" and CC", F'F" and C'C" be drawn in perspective from GKH respectively, we have three pairs of points on each pair of lines in perspective from GKH. Hence the fourth pairs PP', PP", and P'P" must be in perspective from the same three centres. Now let PP" be taken on the common ray KH. Then, since P' must lie on HP", the points PP'P" will fall on the common ray KH. But at the same time G must fall on PP'. Therefore GHK lie on the same line.

DEF.-A transversal is a line drawn across the sides or sides produced of a closed figure or a pencil of rays.

COR. From any pole O (Fig. 2) draw four rays, upon three of which choose any three points abc. Produce ba and bc to intersect any transversal 54 in the points 1 and 2. Next take any point b' on the ray Ob, and join b'l and 6'2 intersecting the rays Oa and Oc in a' and c' respectively. Then by Art. 1 the triangles abc and a'b'c' are correlative, because their corresponding lines meet in the same transversal, and the rays through their corresponding

apices meet in the same point. Next from the apex c

draw any line cd meeting the transversal in the point 3,

and join the points 3 and c' by a line meeting the ray Od in a point d'; then it follows that the triangles acd and d a'c'd', whose sides the same

meet upon

transversal and the

rays through whose

apices meet in the same point, are correlative; wherefore the complete quadrilaterals abcd and a'b'c'd' are correla

the quadrangle klmn (Fig. 3) to meet in the point a, and the sides kn and Im to meet in c. Join ac and produce the diagonals km and In to meet the third diagonal ac in d and b; then the line ad will be harmonically divided in b, c; or, in other terms, the product ad. bc of the whole span ad and the central span be is equal to the product ab. cd of the two end spans ab and cd.

The ratio to be proved is

ab da de + ac

=1+

ас

de;

when, therefore, the point d passes to an infinite distance on the line cd, the length de will be infinite and the ratio

ab

bc

will be equal to unity, which is the case illustrated in

Fig. 4. We shall now proceed to deduce the general

proposition from a consideration of this particular case the formal statement of which is that, if the line fe be drawn parallel to ray Oa and be then bisected in c by the ray Oc, the line fe will be harmonically divided by the point c and the point at infinity, and the points fceco may be regarded as determined by the intersections with line fe of the pairs of opposite sides and diagonals of a quadrangle.

Since cf is parallel to aO, the triangles def and daO are similar, as also are the triangles ceb and abO; wherefore

or the segment db is divided internally at c and externally at a in the same ratio by the harmonic pencil O)fce∞.

be

It remains to proved that the points fceco may be regarded as determined by the intersections with line fe of the opposite pairs of sides and the diagonals of a quadrangle.

Hence in Fig. 5, which is a reproduction of part of Fig. 4, it is necessary to shew that the points FCE may be regarded as determined by the construction of a quadrangle.

CASE I.-Let us take first the simplest case by assuming OC to be per

pendicular to EF. From F draw any indefinite line FG

meeting OE in g and OC in d, and project the point d from E upon the line OF in f. Join and produce fg; then, by construction, the opposite sides gd and Of of the quadrangle Ogdf meet the line FC in a common point F, the opposite sides fd and Og meet the same line in a common point E, the diagonal Od meets FC in C; and, therefore, it only remains to be proved that the diagonal fg produced will meet FC at an infinite distance.

Now in the triangles FdC and EdC, the angles at C are right, the line dC is common, and by hypothesis FC = EC ; hence the triangles are equal in every respect, so that dEdF and angle dEC = angle dFC. Similarly, the triangles FOC and EOC are equal and OE OF, angle OFC OEC; whence by subtraction

=

angle OFd=OEd.

Moreover, the external angle gdE is equal to the sum of the interior and opposite angles dEF and dFE, which sum is also equal to the angle fdF. Thus in the triangles dEg and dFf we have the side dE and the adjacent angles equal to the side dF and the adjacent angles, each to each, wherefore gE=ƒF and

OE-gE or Og=OF-ƒF or Of,

so that

Og Of

and fg is parallel to FE.

OE OF'

CASE II. Let d, f,O,g, be a second position of the quadrangle, where CO, is no longer perpendicular to EF. in Case I. project d, from E upon FO, at ƒ, and join EO, meeting the fixed line FG in g1; then it remains to be proved that fig, is still parallel to FE.

The displacements of the points g and ƒ are

; ffi=dd1E Ff1

where

OE FO1

dE Fdi