Why is split algebraic group quasi-split? One says a connected linear algebraic group $G$ over a field $k$ is quasi-split over $k$ if there exists a Borel subgroup defined over $k$, and is split if there exists a split maximal torus $T$ over $k$.
From those definitions, why is a split algebraic group quasi-split?
 A: $\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\Lie}{\mathrm{Lie}}$Let $\lambda\in X_\ast(T)$ be such that for all roots $\alpha\in\Phi(G,T)$ we have that $\langle \lambda,\alpha\rangle\ne 0$. This is always possible since, essentially, the set of $\lambda$ such that $\langle \lambda,\alpha\rangle=0$ for one $\alpha$ is a $\Z$-hyperplane in $X_\ast(T)$, and a union of finitely many such $\Z$-hyperplane cannot equal all of $X_\ast(T)$. Such a $\lambda$ is called regular. Let $Z_{G}(\lambda)$ be the scheme-theoretic centralizer of $\lambda$. We claim that $Z_{G}(\lambda)=T$. 
Well, $Z_{G}(\lambda)$ is smooth and connected (see Theorem 4.1.7 of Conrad's Reductive Group Schemes notes) and since we have an obvious inclusion $T\subseteq Z_{G}(\lambda)$ it suffices to show that we have an equality of Lie algebras $\Lie(T)=\Lie(Z_{G}(\lambda))$. But, note that this latter Lie algebra can be described in terms of the subalgebra of $\Lie(G)$ where $\lambda$ acts trivially. Note though that for each root space $\mathfrak{g}_\alpha$ we have that $\lambda$ acts on $\mathfrak{g}_\alpha$ by $t^{\langle \lambda,\alpha\rangle}$. By assumption this is non-trivial for all $\alpha$ and thus the claim follows. 
Note then that since $Z_{G}(\lambda)=T$, and thus solvabe, we have that $P_{G}(\lambda)$ is solvable since $P_{G}=Z_{G}(\lambda)\rtimes U_{G}(\lambda)$ (See loc. cit. for the definitions of these objects and this decomposition). But, we know that $P_{G}(\lambda)$ is a connected parabolic subgroup of $G$. Thus, $P_{G}(\lambda)$ being a solvable parabolic subgroup is necessarily a Borel.
EDIT: Another tongue-in-cheek answer is the following:

Theorem(Borel, Borel-Tits???): Let $G$ be a reductive group over a field $k$. Then, $G$ is quasi-split if and only if for any maximal split torus $S$ in $G$ one has that $C_G(S)$ is a maximal torus in $G$.

(note that the above doesn't depend on the choice of maximal split torus since all are conjugate by Grothendieck's theorem)
So, if $G$ is split with maximal torus $T$, then $T$ is also a maximal split torus and since $T$ is self-centralizing we deduce that $G$ is quasi-split! :)
Of course, this is very silly since proving the above statement requires a real deep-dive into the relative root theory of quasi-split groups, which is very complicated. 
