Write $A = B + C$ where $B$ is symmetric and $C$ is antisymmetric.  The assumption is that $B$ is positive-definite and $\det(A) = \det(B)$.  we wish to prove $C=0$.

First, every positive-definite matrix is of the form $g \cdot g^t$ for some $g \in \mathrm{GL}_n(\mathbf{R})$.  Thus, by acting by this group, we can assume $A =1$.

Now, $C$ is anti-symmetric, so by the spectral theorem I can assume $C$ is block diagonal with two-by-two anti-symmetric blocks.  (See the Wikipedia page for "Skew-symmetric matrix").

The determinant is now immediate to compute, and one gets $C=0$.