Let $G$ be a connected affine algebraic group over an algebraically closed field. A parabolic subgroup of $G$ is a subgroup $P$ such that $G/P$ is a complete variety. This is equivalent to $P$ containing some Borel subgroup $B$ of $G$. In Humphreys' Linear Algebraic Groups, exercise 21.10, one has to prove the following:
Let $B$ be a Borel subgroup of $G$. Then $B = N_G(B)^{\circ}$ (and similarly for parabolic subgroups).
For a Borel subgroup, I did the following: $B$ is a normal Borel subgroup of $N_G(B)^{\circ}$, so the variety $N_G(B)^{\circ}/B$ is complete, irreducible and affine, i.e. a point. Thus $B = N_G(B)^{\circ}$. However this does not generalise to parabolics without proving that all parabolics are connected. Is there a way to use this property of Borel subgroups to prove that parabolic subgroups are connected?