1) AA', BB', CC' lines pass all three through a point O (called

2) The intersection points of the sides (prolongations) A''=(BC, B'C'), B''=(CA,C'A'), C''=(AB,A'B') lie on a line e (called

To prove 1) => 2) apply Menelaus (see Menelaus.html ) three times to the triangles:

(a) OAB and line [C''A'B'] => (C''A/C''B)(A'O/A'A)(B'B/B'O) = 1,

(b) OAC and line [B''A'C'] => (B''C/B''A)(C'O/C'C)(A'A/A'O) = 1,

(c) OBC and line [A''C'B'] => (A''B/A''C)(B'O/B'B)(C'C/C'O) = 1.

Multiplying the sides of the equations: (C''A/C''B)(B''C/B''A)(A''B/A''C) = 1, which by Menelaus means that A'', B'', C'' are collinear points on the sides of triangle ABC.

To prove 2) =>1) consider triangles C''BB' and B''C'C which by the current hypothesis are perspective at A''. Hence by the proved part of the theorem corresponding side-pairs intersections O= (BB', CC'), A=(BC'', CB'') and A'= (B''C', C''B') are on a line, proving that ABC and A'B'C' are point perspective with respect to O.

See the file Duality.html for another kind of proof of 2) => 1) based on the duality of projective spaces.

According to the corollary all triangles A'B'C' having their vertices {B',C'} on lines {OX,OY} respectively and their sides passing through the fixed points {A'',B'',C''} have their third vertex A' varying on a fixed line OA. Its intersection with the third side XY of the given triangle determines point A' of the desired triangle. This determines also completely triangle A'B'C', since its sides must pass through the given points {A'',B'',C''}.

When a polygon varies so that its n sides pass through n fixed points lying on a given line and (n-1) of its vertices lie on (n-1) given fixed lines, then its n-th vertex describes also a line.

To prove this consider the intersection points {V,W} of line-pairs correspondingly (AB'',OB) and (A''B'',C''B). Use a projective base in which A''(1,0,0), B''(0,1,0), C''(0,0,1) and calculate the coordinates of A:

Let lines {OX, OY, A''BC} be respectively of the form

OX: ax+by+cz=0,

OY: a'x+b'y+c'z=0,

A''BC: uy+vz=0.

Points {B,C} have then coordinates

(bv-cu, -av, au),

(b'v-c'u, -a'v, a'u).

Lines {BC'',CB''} have respectively the form:

(-av)x + (cu-bv)y=0,

(-a'u)x + (b'v-c'u)z=0.

Their intersection is point A with coordinates:

A: ((cu-bv)(b'v-c'u), av(b'v-c'u), a'u(cu-bv)).

Since lines {OX,OX} intersect at a point of line x=0, we find that b'=sb, c'=sc for a constant s.

Thus, the coordinates of point A become:

(-s(cu-bv)

On the other side the coordinates of points {S,R} are easily calculated to be

S: (-b,a,0) and R: (-sc,0,a'). The determinant of the coordinates of these two and (*) is then:

This proves the collinearity of points {A,S,R}. By Desargues theorem triangles A''WC'' and OCV, being line-perspective, are also point-perspective with respect to a point U. One could start from this and show the collinearity of three points {A,S,R}. The calculation though seems to be slightly more complicated than the one given above.

PerspectivityAndPerspectiveTriangles.html

Harmonic_Perspectivity.html

Duality.html

Maclaurin.html

Menelaus.html

Desargues_Many.html

MenelausApp.html

PappusLines.html

PappusSelfDual.html

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