Let $ABC$ be a triangle where $AB > BC$, and $D$ and $E$ be points on sides $AB$ and $AC$ respectively, such that $DE$ and $AC$ are parallel. Consider the circumscribed circumference of triangle $ABC$. A circumference that passes through points $D$ and $E$ is tangent to the arc $AC$ that does not contain $B$ at point $P$. Let $Q$ be the reflection of point $P$ with respect to the perpendicular bisector of $AC$. The segments $BQ$ and $DE$ intersect at $X$. Prove that $AX = XC$.