I assume $ABCD$ is circumscribed. Stewart's theorem on $ABC$ (or the median formula) shows that $BX^2 = AB\cdot BC$: in fact, $BX^2=\frac14(2AB^2+2BC^2-AC^2) = \frac14(2AB^2+2BC^2-2(BC-BA)^2)=AB\cdot BC.$ Relations like $BX^2 = BA\cdot BC$ reminds us of power of a point. Indeed, consider point $A'$ on side $BC$ such that $BA'=BA$. Then $BA'\cdot BC=BX^2$ shows that $BX$ is tangent to the circumcircle of $CXA'$. Then $\angle A'XB = \angle BCA$. We also have $CA' = BC-BA=AC/\sqrt2$, so $CX\cdot CA = AC^2/2=CA'^2$, and $CA'$ is tangent to the circumcircle of $A'XA$, and $\angle A'XA = \angle BA'A = 90^\circ - \frac12\angle ABC$. Thus $\angle BXA = \angle A'XA - \angle A'XB = 90^\circ - \frac12\angle ABC - \angle BCA = \frac12(\angle BAC - \angle BAC)$. Noting that $CD-AD=BC-BA$, we can see that triangle $ADC$ also has $\sqrt2(CD-AD)=AC$, so $\angle DXA = \frac12(\angle DAC - \angle DCA)$. Summing these two inequalities yield the result.

As above, derive that $BX^2 = BA\cdot BC$. Now the $\sqrt{ac}$ inversion centered at $B$ sends $X$ to the intersection $D$ of the $B$-symmedian with $\odot(ABC)$. The metric condition then implies $BX = BD$. To finish, define $B'$ as the reflection of $B$ over the perpendicular bisector of $\overline{AC}$. Then $B$, $X$, and $D$ are collinear by symmetry, and furthermore $XA$ bisects $\angle BXD$ (chord-chord angle formula). Now angle chase to get \[ 2\angle BXA = \angle BXD = \angle BDX = \angle A - \angle C. \]