[2] All trilinear polars tr(E) of points E lying on such a conic are parallel to each other and define a point at infinity P

[3] P

[4] The trilinear polar L=tr(P

[5] Line L is also the isotomic dual of the tangent t

[6] The isotomic t(P

[7] The line through {G,t(P

[8] The parallelogram with vertices {t(P

[9] Inversely every conic having its perspector P

The proof is trivial in barycentric coordinates. In fact, the conic through {A,B,C} has an equation of the form

a/x + b/y + c/z = 0.

If G(1,1,1) is on the conic then a+b+c=0, which means that P

On the other side the trilinear polar of a point X(u,v,w) on the conic is

x/u + y/v + z/w = 0.

But a/u + b/v + c/w = 0 implies that this line passes through P

That this conic is actually a hyperbola follows fromt the fact that it is the isotomic image of line ax+by+cz=0 which passes through G. Thus it intersects the outer Steiner ellipse at two distinct points, which by isotomy map to points at infinity. This shows [1].

[4] Is a consequence of the discussion in HyperbolaPropertyParallels.html where the construction of the trilinear polar of a point at infinity is carried out. To find the trilinear polar L=tr(P

(i) Draw parallels to tr(E) from the vertices of ABC.

(ii) Find the intersections {A',B',C'} of these parallels with the opposite sides of ABC.

(iii) Take the harmonic conjugates {A'',B'',C''} of the former points with respect to {(B,C),(C,A),(A,B)} respectively.

Line L=tr(P

Since L is the polar of the point at infinity P

[5] follows from a trivial calculation showing that line ax+by+cz=0, containing G(1,1,1), is tangent at to the conic at G. Since this line maps through isotomy to the conic it is the dual of the perspectrix tr(P

To show property [6] use the previously proved t

The matrix is the one expressing the conic as a quadratic form. The claim in [6] follows from this matrix identity.

To prove [7] it suffices to express G'=t(P

q(G',G')=0, q(X,Y) denoting the quadratic form expressing the conic through the matrix considered above.

Replacing the equation one calculates easily that s = (a

Express now G in terms of G' and t(P

A short calculation reduces this to showing that a

To show [8] let {S

Hence lines S

These lines pass also through t(P

To show this consider the barycentric coordinates of S

ax+by+cz=0. (1)

Since it is also on the outer Steiner ellipse it satisfies (1/x) + (1/y) + (1/z) = 0. (2)

In addition we have a+b+c = 0. (3)

The collinearity stated for line S

Solving (1) and (3) with respect to (a,b,c) we find:

(a,b,c) = ( y-z , z-x , x-y ). (4)

Introducing this into the above determinant and expanding it we get the equivalent equation

Last equation is a consequence of (2). This proves the claim for line S

To show [9] consider the tripols of lines passing through P

Desargues.html

Harmonic.html

Harmonic_Bundle.html

HyperbolaPropertyParallels.html

IsotomicConicOfLine.html

LineInTrilinears.html

TriangleConics.html

Trilinears.html

TrilinearPolar.html

Tripole.html

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