Proof
For F = f:g:h, let F* be the point p"(g/q'+h/r'):q"(h/r'+f/p'):r"(f/p'+g/q').
If F is on C, so that p'gh+q'hf+r'fg=0, then a simple calculation shows that F* is on the inconic
I : (x/p")2+(y/q")2+(z/r")2-2yz/q"r"-2zx/r"p"-2xy/p"q"=0.
Another simple calculation shows that the tangent to I at F* has equation fx/p'p"+gy/q'q"+hz/r'r"=0.
Now suppose that U = u:v:w and K = k:l:m, U ≠ K, lie on C.
Note that U,K lie on the tangent at F* if and only if
fu/p'p"+gv/q'q"+hw/r'r"=0, and
fk/p'p"+gl/q'q"+hm/r'r"=0.
As U ≠ K, these as equations in f,g,h have a solution unique up to scaling.
As U, K are on C, it is easy to check that f = p'2p"/uk, g = q'2q"/vl, h = r'2r"/wm is a solution.
If these hold, then, by symmetry, F lies on the tangents at U* and K*.
Thus ΔFUK is inscribed in C, and touches I at F*,U*,K*.
Theorem TC10'
Suppose that C and I are the circumconic and inconic with perspector P = p:q:r.
For F = f:g:h on C, the triangle with vertex F inscribed in C and touching I has other vertices
U = pg/q:qh/r:rf/p, K = ph/r:qf/p:rg/q (and F* is the "barycentric sum" of U and K).
Proof
As F = f:g:h is on C, pgh+qhf+rfg = p/f+q/g+r/h = 0.
As in the above proof, the tangent at F* is the line L :
fx/p2+gy/q2+hz/r2=0.
Then it is easy to check that the given points U and K are on C and L.