I am currently in Chapter 12 of 'Lemmas in Olympiad Geometry' by Titu Andreescu et al. They have stated and given a satisfactory proof Brianchon's Theorem using poles and polars as well as Pascal's theorem. In the end, the author mentions:
Brianchon's Theorem is what's known as a projective dual of Pascal's
theorem. Consider the 2 statements: every 2 distinct points determine
a line and every 2 distinct lines determine a point(possibly at
infinity); These are projective duals. A projective dual can be
obtained by switching every pole with its polar and vice versa.
How can we switch the poles and polars of Pascal's to get Brianchon's, or of Ceva's to get Menelaus'?
Best Answer
I think part of the confusion is that the author seems to be conflating the concepts of polar duals and projective duals. These points are not the same and to define the former you need to have non-degenerate conic around. Let me try to explain the difference. In polar duality there is a correspondence between lines and points in the same space. In projective duality there is a correspondence between lines and points in different spaces. More precisely, lines in projective space correspond to points in the dual projective space. This is more or less tautological since we can define dual projective space as the space of lines in projective space if we like. To get the correspondence between points in projective space and lines in the dual projective space we look at the set of all projective lines through our point. This turns out to form a line in the dual projective space.
In the case of Pascal and Brianchon's theorems there is a conic lying around in the statements of the theorems and so it makes sense to say that one theorem is the polar dual of the other (I will come back to this in a bit). For the theorems of Ceva and Menelaus there is not conic lying around and so it doesn't make sense to say that one is the polar dual of the other, but it does make sense to say that one is the projective dual of the other and this ends up being true (again, more about this soon).
To see that Brianchon is the polar dual of Pascal, let's look at the picture below. In what follows whenever I decorate an object with $\perp$ sign this means to take the polar dual with respect to the conic in the picture.
From Pascal's theorem we have that $s_1\cap r_1$, $s_2\cap r_2$, $s_3\cap r_3$ are all in $\ell$. Thus we have that $\ell^\perp\subset (s_1\cap r_1)^\perp\cap(s_2\cap r_2)^\perp\cap(s_3\cap r_3)^\perp $. These will end up being the three lines intersecting at a point that show up in the conclusion of Brianchon's theorem, we just need to find the hexagon. The points $r_1$, $r_2$, $r_3$, $s_1$, $s_2$, and $s_3$ all lie on the conic and so their polar duals are 6 lines tangent to the conic that form the sides of a hexagon with vertices $r_1^\perp\cap r_2^\perp$, $r_2^\perp \cap r_3\perp$, $r_3^\perp\cap s_1^\perp \ldots s_3^\perp\cap r_1^\perp$. One thing that makes this all confusing is that the standard pictures are misleading in the sense that if you take the polar dual of the picture above you don't get the standard Brianchon picture where the intersection point of the three lines appears inside the conic. This is easy to see since the intersection point is the polar dual of the line $\ell$ which intersects the inside of the conic and so its polar dual will be a point on the outside of the conic.
As for Menelaus and Ceva, there is a nice way to view the projective duality between them as well as providing a nice proof. Let $P$ be real projective 2 space (i.e. non-zero vectors in $\mathbb{R}^3$ up to scaling) and $P^\ast$ the dual projective space (non-zero linear functionals up to scaling). One way to think about $P^\ast$ is as lines in $P$, with the correspondence being given by taking a linear functional to its kernel.
A flag in $P$ consists of a pair $(p,l)$ of a point and a line such that $p\in l$. Given 3 flags in $P$ $((p_1,l_1),(p_2,l_2),(p_3,l_3))$ we can define their triple ratio by
$$ \frac{l_1(p_2)l_2(p_3)l_3(p_1)}{l_1(p_3)l_2(p_1)l_3(p_2)} $$
where when we evaluate $l_i(p_j)$ we are picking representatives of $p_j$ and $l_i$ in $\mathbb{R}^3$ and its dual and evaluating $l_i$ at $p_j$. This is well defined since each $p_i$ and $l_j$ occurs exactly once in the numerator and the denominator and so the ambiguity of our choice cancels out. It is not hard to check that the triple ratio of a triple of flags is unchanged by applying a projective transformation to the 3 flags.
This construction may seem strange at first, but if you think about it a bit then you will realize that it is basically the 6 term expression that appears in the statement of Menelaus and Ceva's theorems. For instance if we pick the correct scaling for $l_1$ then its value computes distance from $l_1\cap p_2p_3$ along the line $p_2p_3$.
Assuming the flags are in sufficiently generic position, we can apply a projective transformation so that $p_1=[e_1]$, $p_2=[e_2]$, $l_2=[e_2^\ast]$, $l_2=[e_1^\ast]$, and $p_3=[e_1+e_2-e_3]$, where $\{e_1,e_2,e_3\}$ is the standard basis on $\mathbb{R}^3$. In these coordinates we can then assume that $l_3=[ae_1^\ast+e_2^\ast+be_3^\ast]$, however, since we need $p_3\in l_3$ it follows that $b=1+a$. In these coordinates the triple ratio is equal to $l_3(p_1)=a$ (we have rigged things so that the other terms are all equal to 1). See the picture below
Now let's think about what it means for the triple ratio to equal -1. This means that $l_3=[-e_1^\ast+e_2^\ast]$. In our coordinates we have that $l_1\cap l_2=[e_3]$, and so the triple ratio equals -1 if and only if $l_1$, $l_2$, and $l_3$ have a common point of intersection ($[e_3]$ in these coordinates). This is just a rephrasing of Ceva's Theorem.
For Menelaus' theorem we use a slightly different coordinate system. This time around we let $p_1=[e_1]$, $p_2=[e_2]$, $l_1=[e_2^\ast]$ and $l_2=[e_1^\ast]$ and $l_3=[e_1^\ast+e_2^\ast-e_3^\ast]$. With this normalization $p_3=[e_1+ae_2+(1+a)e_3]$ (i.e. we have normalized the third line rather than the third point). In this coordinate system we see that the triple ratio is again equal to $a$. Thinking again about what it means for the triple ratio to be -1 we find that this happens if and only if $p_3=[e_1-e_2]$. However $[e_1-e_2]$ is the intersection between $p_1p_2$ and $l_3$ and so the triple ratio is -1 if and only if $p_3$ lies on $p_1p_2$. This is Menelaus' theorem.
One way of putting this all into context is that both Ceva and Menelaus' theorems are describing the triple ratios of triples of flags that are degenerate. In Ceva the configuration is degenerate because the three lines intersect and in Menelaus it is because the 3 points live on a line. These two conditions are dual to one another in the sense that three points in the dual space $P^\ast$ lie on a line in $P^\ast$ if and only if the corresponding three lines intersect. This is the sense in which Menelaus and Ceva are dual to one another.