In $\triangle A B C$, with usual notations, if $\cos \frac{B}{2}=\sqrt{\frac{c+a}{2 a}}$, then $a^2=$

In triangle $\triangle ABC$, the given relation is $\cos \frac{B}{2} = \sqrt{\frac{c+a}{2a}}$.

Squaring both sides yields $\cos^2 \frac{B}{2} = \frac{c+a}{2a}$.

Using the identity $\cos B = 2 \cos^2 \frac{B}{2} - 1$, we substitute to obtain $2 \left(\frac{c+a}{2a}\right) = 1 + \cos B$, which simplifies to $\frac{c+a}{a} = 1 + \cos B$.

Solving for $\cos B$ gives $\cos B = \frac{c+a}{a} - 1 = \frac{c}{a}$.

Applying the Law of Cosines at angle $B$: $b^2 = a^2 + c^2 - 2ac \cos B$. Substituting $\cos B = \frac{c}{a}$ produces $b^2 = a^2 + c^2 - 2ac \cdot \frac{c}{a} = a^2 + c^2 - 2c^2 = a^2 - c^2$.

Rearranging gives $a^2 = b^2 + c^2$, which corresponds to option C.

The final answer is $\boxed{\text{C}}$.