Let $A, B, C, a, b, c$ be three points and three lines respectively.
Prove (or provide a counterexample) that:
$\triangle ABC$ and $\triangle abc$ are perspective if and only if there exists a conic $\Omega$ (possibly imaginary) such that $(A,a),(B,b),(C,c) $ are pole-polar pairs with respect to $\Omega$.
The reverse proof isn't that important as I already have proof for it using Sondat's Theorem but feel free to post it.
Reverse proof:
Perform a projective transformation taking the conic to a circle. Let its center be $O $. From the definition of the polars $\triangle ABC $ and $\triangle abc$ are orthogonal in both directions from $O $ and thus by Sondat's Theorem they are perspective.$\blacksquare$
Partial forwards proof:
Let $a \cap \overline{CB}, \overline{AC}, \overline{AB}, c, b = F, Y, X, B_1, C_1, b \cap \overline{AB}, \overline{AC} = P, E, c \cap \overline{AB}, \overline{AC} = D, Q$.
Suppose $(X,C_1;Y,B_1) < 0$. Let $O $ be such $\measuredangle XOC_1 = \measuredangle YOB_1 = 90^\circ$($O$ exists because of the assumption).
Perform a projective tansformation that takes $X, Y, B_1, C_1$ to $\overline{OX}_\infty, \overline{OY}_\infty, \overline{OB_1}_\infty, \overline{OC_1}_\infty$.
Now work on the resulting picture.
Since the triangles are perspective we have $\overline{DEF}$. Since $a $ was taken to infinity, $F $ was taken to infinity and thus $\overline{BC} \parallel \overline{DE}$.
From the projection we have $X \perp C_1, Y \perp B_1$ thus $b \perp \overline{AB}, c \perp \overline{AC}$ thus $\overline{PQ}$ is antiparallel to $\overline{DE}$ within $\measuredangle BAC$ thus $\overline{PQ}$ is antiparallel to $\overline{BC}$ within $\measuredangle BAC$ thus $AP \cdot AB = AQ \cdot AC $.
Now if $AP \cdot AB $ is positive draw a circle of radius $\sqrt{AP \cdot AB}$. If it's negative draw an imaginary circle. Thus there exists a circle such that $(A,a), (B,c), (C, c) $ are pole-pairs with respect to it. Reversing the transformation yields a conic in the original picture. $\blacksquare$
