Proving that central simple algebra $B$ over a field $k$ splits in Brauer group by a field $L$ contained in $B$, then $[B:k]=[L:k]^2$

Question

I want to prove the following statement (I read it here - https://www.jmilne.org/math/CourseNotes/CFT.pdf#X.4.3.6)
The similarity in the below statement comes from the equivalence relation from the Brauer group.

Let $A$ be a central simple algebra over $k$. A field $L$ of finite degree over $k$ splits $A$ if and only if there exists an algebra $B$ similar to $A$ containing $L$ such that $[B:k]=[L:k]^2 $

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The proof is clear to me except how we get in proving forward implication $[B:k]=[L:k]^2 $.
I know this is equivalent to saying that $L$ is its centralizer in $B$ or $L$ is maximal commutative $k$-subalgebra inside $A$.

But I do not understand how to go about proving this from here.

Also does this imply that $L$ splits $B$ ?

Answer 1

Yes, $L$ does split $B$ (follows from the last line of the corollary in question).
From the first line in proof $[A:k][L:k]=dimV_L^2 [L:k] $ and second last line of first paragraph $[B:k][A:k]=dimV_L^2[L:k]^2 $. From these two equations the desired equality follows.