Théorème de Wederburn I ask for Wedderburn's Theorem.
my question is if $R$ is a simple ring, then $R = M _{n}(D)$ by the Theorem of Wedderburn ?
and how can conclude that  $ R=D $ or $R = M _{2}(D)$ by the following lemma:
If L is a proper left ideal of R then L is both minimal and
maximal.
reference:

Bergen, Jeffrey, I. N. Herstein, and C. Lanski. Derivations with invertible values. Canad. J. Math 35.2 (1983): 300-310.

 A: 
if R is a simple ring. then $R = M _{n}(D)$ by the Theorem of Wedderburn ?

No. The theorem you are thinking of only applies to simple Artinian rings.
There are lots of simple rings that aren't Artinian, and they are not isomorphic to such matrix rings.
If you have any further doubts, check this question:
why is a simple ring not semisimple?
Also related is: Definition of a simple ring .
Perhaps the context of the paper includes Artinianness (or maybe works with finite dimensional algebras, which are Artinian), but I can't access the paper. Further details would be helpful.

how can conclude that  $ R=D $ or $R = M _{2}(D)$ by the following lemma:
  If L is a proper left ideal of R then L is both minimal and
  maximal.

I don't think this lemma is stated correctly or completely. The only proper left ideal of a division ring $D$ is the zero ideal, which isn't a minimal left ideal.  (Of course if you allowed the zero ideal to be minimal, then $M_2(D)$ would only have one minimal left ideal, and it wouldn't be maximal.)
Even overlooking that point, the statement is apparently incorrect as written: $\mathbb R[x]/(x^2)$ has exactly one proper ideal $(x)/(x^2)$, and it is both minimal and maximal, and this ring isn't semisimple because it has nonzero nilpotent elements. So, something is amiss.
