A nice example is Pascal's theorem for the circle:
If a hexagon is inscribed in a circle then the intersections of opposite sides are collinear.
Pluecker
Plücker gave an elegant proof of Pascal's theorem as a consequence of Bezout's Bézout's theorem. View the unions of alternate sides of the hexagon as cubic curves
$l_{135}=0$ and $l_{246}=0$.
They meet in 9 points, 6 of which are the vertices on the circle $c=0$. But we can choose constants $a,b$ so that the cubic
$al_{135}+bl_{246}=0$
passes through any point. Taking this point on the circle, the circle and the cubic have at least 7 points in common. By Bezout's Bézout's theorem, the curves have a common component, necessarily the circle $c=0$, since $c$ is irreducible.
Hence $al_{135}+bl_{246}=cp$, for some polynomial $p$, which must be linear. Since $al_{135}+bl_{246}=0$ contains all 9 points common to $l_{135}=0$ and $l_{246}=0$, while $c=0$ contains only 6, the remaining 3 (intersections of opposite sides of the hexagon) must lie on the line $p=0$.
