If $K / F$ is an algebraic extension, then $\overline{K}$ is an algebraic closure of $F$.

In particular, $\overline{F} \cong \overline{K}$. The converse is not true, i.e. $\overline{K} \cong \overline{F}$ does not imply $K / F$ is algebraic, because it can happen $K / F$ is not algebraic and $K \cong F$, for example, $F = \mathbb{Q}(x_{1}, x_{2}, \dots, x_{m}, \dots), K = \mathbb{Q}(x_{0}, x_{1}, \dots, x_{n}, \dots)$.