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We can rephrase the problem as follows: Let $P$ be the incenter of triangle $XAD$, and let $Q$ be a point on $AD$. Construct points $B$ and $C$ on $AX$ and $DX$, respectively, such that triangles $QAB$ and $QDC$ are isosceles. Show that $PB=PC$. Stated this way, this problem is actually very easy since immediately we have that $PB=PQ=PC$ since $P$ lies on the perpendicular bisectors of $BQ$ and $CQ$. In fact, $P$ is the circumcenter of triangle $BQC$.