The proof is modeled as an exercise in projective coordinates with respect to the fixed points {A*,B*,C*} and with coordinator the fixed point A

y-kz=0 (with variable k),

and lines {A

x-ay-bz=0,

x-a'y-b'z=0. with fixed (a,b), (a',b') satisfying 1=a+b=a'+b' (*), since A

The coordinates of {B,C} are correspondingly:

(ka+b, k, 1) and (ka'+b', k, 1).

Then lines {BC*,CB*} obtain the form:

-kx+(ka+b)y=0 and -x+(ka'+b')z=0.

And eliminating k from them leads to equation:

ba'yz = (b'z-x)(ay-x) (**)

representing a conic passing through B*(0,1,0), C*(0,0,1) and A

To prove the inverse one can use Maclaurin's construction to define a conic (c') with the given data. This conic then will have with (c) the common points {A

meet on a line (BC). Maclaurin's construction in this case shows that the quadrilaterals A

ConicsMaclaurin.html

ConicsMaclaurin2.html

Salmon, G.

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