AoPS - Problem

Sketch of my solution: Let $Z'$ be point on $\Gamma$ such that $BD$ is tangent to $ADZ'$. We try to show that $BZ'$ cuts $AX$ at its midpoint. $AE$ intersects $BC$ at $L$. It is obvious that $(D,L,C,B)=-1$ and $\angle CZB$=$90$, $\angle CAB=90$. So, line $AC$ and $ZC$ are angle bisectors of $\angle DAL$ and $\angle DZL$, respectively. After easy angle chasing we get that $DZ$ is also tangent to $(AZL)$. So, $AZ$ intersects $DL$ at its midpoint. $AX$ intersects $BZ'$ at $Y'$. After applying sine law to triangle $ADL$ , we get that $\frac{\sin \angle DAZ}{\sin \angle ZAL}$=$\frac {\sin \angle ADL}{\sin \angle ALD}$. $\angle DAC=\angle Y'BA$ , $\angle ZAL=\angle Y'BE$, $\angle ZAC=\angle ZBC$, $\angle LDA=\angle XAB$. So, we get that $\frac{\sin \angle Y'BA}{\sin \angle Y'BE}$=$\frac {\sin \angle XAB}{\sin \angle AXB}$. Which implies $Y'$ is the midpoint of $AX$.