> I don't know if the following theorem is known
> Let A'B'C' be the cevian triangle of a point P wrt a triangle ABC.
> B'C' intersects the circumcircle of ABC at A1 and A2.
> It is not important on which side of BC you choose A1 or A2.
> Define B1, B2, C1 and C2 similarly.
> The lines A1B1, A2C1 and B2C2 form a triangle A*B*C* perspective
> with A'B'C'
> In fact there are 8 possible triangles A*B*C* by changing A1 with
> A2, B1 with B2 and C1 with C2
> However they only lead to 4 different perspectors
This is true with any circle, or even conic. The theorem follows from
Pascal's hexagon theorem, combined with Desargues' two triangle theorem.
See for an explanation (also why there are only 4 perspectors) my paper
Equilateral Chordal Triangles in FG 2, 33-37.
> What is the area of the plane so that for a point inside this area
> all three sides of its cevian triangle intersect (or at least touch)
> the circumcircle ?
With P(x:y:z) we find the quartic
a^4y^2z^2 + b^4x^2z^2 + c^4x^2y^2
- 2a^2b^2xyz^2 + 2a^2c^2xy^2z + 2b^2c^2x^2yz = 0
to let the line A'B' be tangent to the circumcircle.
Similar quartics for the other Cevians. The point P has to be inside all
three of them.