Incentric Ratio

$\angle CEI = 90 \Rightarrow O$ is a midpoint of $CI$ where $O$ is circumcentre of $\triangle CEI$ .

Let $Z$ be the midpoint of $AC$ . Then $OZ \parallel IA$ . $IA \perp FE \Rightarrow OZ \perp FE$ . Let $G$ be a point of intersection of the extensions of $FE$ and $OZ$ . Then $\triangle EOP$ is isosceles and $G$ is a midpoint of $EP$ . Triangles $AFE, EPZ$ are similar therefore $PZ \parallel AB$ therefore $X$ is a midpoint of $BC$ .