Given $\triangle ABC$ and $AF \bot BC\ ,\ BE \bot AC$. $AF \cap BE=H$, circumcircle (in blue) through $A,B,H$, point $P$ is lying on this circle. $AP \cap BC =A'$ and $BP \cap AC =B'$, midpoint of $A'B'$ is the point $D$. Because $\angle A'AF = \angle PAH=\angle PBH = \angle B'BE$, the triangles $A'AF$ and $B'BE$ move in the same way. If $P \equiv A$, then $D \equiv E$. $x_{C} > \frac{x_{B}}{2}$, if $P \equiv B$, then $D \equiv G$, point $G = EF \cap BI$, point $I$ is constructed $\angle BAF = \angle EBI$. Locus is the line segment $EG$ (in red).

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