Let $ABCD$ be a cyclic quadrilateral and apply Pascal's Theorem to $AABCCD$ and $ABBCDD$. Then this is just the Brocard's Theorem.
This is my approach:
Let the tangents on $A$ and $C$ meet at point $P$, and the tangents on $B$ and $D$ meet at point $T$. Then let $Q$ be the intersection point of lines $AB$ and $CD$, and let $R$ be the intersection point of lines $AD$ and $BC$. By the Pascal's Theorem on $AABCCD$,we have that points $P$,$Q$,$R$ are collinear, then again by the Pascal's Theorem on $ABBCDD$ we also have that points $Q$,$R$,$T$ are collinear. Therefore points $P$,$Q$,$R$, and $T$ are collinear, but ...
How does this lead to the Brocard's Theorem?
Can anyone help explain this? Thanks.
Notice that since $PA$ and $PC$ are tangent at the circle, the intersection $X:= AC\cap BD$ lies on the polar of $P$, which is clearly $AC$. Similarly, $X$ lies on the polar of $T$. But that means that $P$ must lie on the polar of $X$ and $T$ must lie on the polar of $X$, so $PT$ is the polar of $X$. But since $P,Q,R,T$ are collinear, $Q$ and $R$ are also on the polar of $X$ so $RQ$ is the polar of $X$. The others follow a bit differently, but they are not to difficult either.