Scalar extension of simple algebras

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I have a question regarding quaternion algebras. Let $K$ be a number field and $L|K$ a quadratic field extension. Let $M|K$ be a field extension such that $M\otimes_K L$ is not a field, i.e. $M\otimes_K L \cong M\oplus M$. Now I consider a central simple algebra $\mathcal{D}$ over $L$, i.e. the center of $\mathcal{D}$ is $L$. If necessary, we can restrict ourselves to the case of $\mathcal{D}$ being a quaternion algebra over $L$. Is it true in general that $A:=\mathcal{D}\otimes _K M$ is a direct sum of two isomorphic central simple algebras over $M$ and if so, why is that true?

We certainly have $Z(A) \cong Z(\mathcal{D}) \otimes_K M \cong M\oplus M$ (where $Z(\phantom{x})$ denotes the center) and that $A$ is a semisimple $M$-algebra. So by Wedderburn's theorem it is a direct sum of two matrix rings over central $M$-division algebras $D_1$ and $D_2$. I.e. $A\cong D_1 ^{n_1\times n_1}\oplus D_2^{n_2\times n_2}$. Is it true that $D_1\cong D_2$ and $n_1=n_2$?

I came across this problem in a much more specific context, but my gut tells me that the claimed statement may be true in the context provided here...

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The answer is yes. One has to check that the isomorphism $L\otimes_K M \cong M\oplus M$ is $L$-linear, so that it is an isomorphism of $L$-algebras. To do so, it is necessary to notice that the condition "$M\otimes_K L$ is not a field" implies that $L$ is a subfield of $M$.

Then one has $$D\otimes_K M \cong (D\otimes_L L) \otimes_K M \cong D\otimes _L (M\oplus M) \cong D\otimes_LM \oplus D\otimes_L M. $$ Since $D$ is a central simple $L$-algebra and $M$ is a simple $L$-algebra, we have that $D\otimes_L M$ is simple, so it is isomorphic to $\tilde{D}^{k\times k}$ for some $k\in\mathbb{N}$ and some central $M$-division algebra $\tilde{D}$.

$D$ being central simple over $L$ is the only necessary condition on $D$.

This generalizes to the situation where $L$ is an arbitrary finite extension of $K$ (not necessarily of degree $2$) such that $M\otimes_K L \cong \bigoplus_{i=1}^{[L:K]} M$.