Finite dimensional central division $\mathbb K$-algebra as a subalgebra of a matrix $\mathbb K$-algebra The question is as follows:


A finite-dimensional central division $\mathbb K$-algebra $D$  is a $\mathbb K$-algebra isomorphic to a subalgebra  of $M_r(\mathbb K)$ if and only if $\dim_{\mathbb K} D \mid r$.


In what comes, we already know that the other direction is true by a beautiful answer from @Algeboy. 
Also, I understood that the comment by @Jyrki give a very nice and reasonable way for to show the statement. But I don't know how to use it.
For to prove this statement, I think for to show the statement, I think we need to use the notion of splitting field for the central division algebra $D$
of dimension say $d^2$ and to show that $[L:K]=rd$ by using $[D:K]=[L:K]^2$. I know how to prove these facts but I still do not know how to apply it for to prove our statement?
Also there is some hint for to prove it in the book Finite dimensional division algebras over fields (1996 Springer) by Nathan Jacobson and you can download it by clicking on this link, in the beginning of the section 2.12 is written that this is a direct sequence from the theorem 2.3.17 and some statements from its proof. But I still cannot see it!
Thanks!
 A: Write $d=\dim_K(D)$. Consider the set $D^m$ for a natural number $m$.
Then $D^m$ is a right $D$-module, and has dimension $md$ as a vector
space over $D$. Choose a basis $B$ for $D^m$ over $K$. Each right multiplication map $\rho_a:x\to xa$ on $D^m$ is a $K$-vector endomorphism
of $D^m$. Expressing it in terms of the basis $B$ gives a matrix $N_a$.
Then $a\to N_a$ is a homomorphism from $D$ to $M_{md}(K)$. As $D$
is simple, its image is isomorphic to $D$ and is a subring of $M_{md}(K)$.
Conversely consider an embedding $\phi:D\to M_n(K)$. Then $K^n$
becomes a right $D$-module via $v\cdot a=v\phi(a)$. For a division
ring, module theory follows the same lines as vector space theory
over a field: every module is free. Thus $K^n\cong D^m$ as a $D$-module
for some $m$. Then $n=md$ on counting dimensions over $K$.
A: If $D$ is a unital subring of $M_r(K)$ then $K^r$ is a unital $D$-module.  But $D$ is a division ring so this means $K^r$ is a $D$-vector spaces, i.e. $K^r=D^m$ for $m=\dim_D K^r$.  Evidently $r=m\dim D$.
