The $ABC$ acute triangle's orthic triangle is the $DEF$ triangle, $D\in AB, E\in AC, F\in BC$. Let $F'$ and $F"$ be points on $DE$, such that $EF'=EF, DF"=DF$ and neither $F'$ nor $F"$ lie inside the circumcircle of triangle $ABC$. Let the center of the circumcircle $AF'F"$ be point $O$. Furthermore let $O'$ be the circumcenter of trianle $ABC$, $e$ be the line tangent to the circumcircle of $\triangle ABC$ at point $A$ and $X\neq A$ be the intersection of $AB$ line and the circumcircle of $AF'F"$. Prove that lines $e, DE, O'B, OX$ define a paralelogram.
Well, I proved that $F'$ and $F"$ is the reflection of $F$ to $AB, AC$. Note that the $AF'F"$ is an isosceles triangle with $\angle F'AF" = 2\times \angle BAC.$ Maybe I should join the perpendicular bisector of $F'F"$? How can I prove it?