Wedderburn-Artin's theorem:

A ring is left semisimple ring iff it is finite product of $$M_{n_i}(D_i)$$ for some division ring $$D_i$$.

Due to this theorem,we know that

if a ring is left semisimple,then it is also right semisimple.

I wonder how to prove this result without using the theorem.the result is equivalent to the following:

if $$l.gl.dim(R)=0$$,then $$r.gl.dim(R)=0$$.

is there a direct proof?

I'm not sure this is really more direct than using Wedderburn-Artin, but a condition on a ring $$R$$ that is clearly left-right symmetric and is equivalent to semisimplicity is:
There is a decomposition $$1=e_1+\dots+e_n$$ into orthogonal idempotents such that, for every $$i$$, $$e_iRe_i$$ is a skew field and, for every $$i$$ and $$j$$, either $$e_i$$ and $$e_j$$ are conjugate or $$e_iRe_j=0$$.