Nice Problem! Define $E$ as the $C$-foot and $G$ as the point such that $ABCG$ is a parallelogram. Then note that since $\angle ABM = \angle XAM$, it follows that $\triangle BAM \sim \triangle AXM$, so it follows that $MX \cdot MB = MA^2$ which implies that $AXCG$ is cyclic. Likewise, $\measuredangle DAG = \measuredangle (AD, AG) = \measuredangle (AD, BC) = 90^\circ$ and $\measuredangle DCG = \measuredangle (DC, AB) = 90^\circ$, so pentagon $AXDCG$ is cyclic. As such, it follows that $\measuredangle EBX = \measuredangle MAX = \measuredangle CAX = \measuredangle CDX = \measuredangle EDX$ so $(EBDX)$ is cyclic which implies that $\measuredangle BXD = \measuredangle BED = 90^\circ$.

Another approach: Note that $\triangle MAX \sim \triangle MBA$ and thus $MX \cdot MB = MA^2$. This makes $X$ the so called $B$-HM point in $\triangle ABC$, which promptly implies the conclusion.

Note that $X$ lies on the $B$-median as well as the circle through $B$ tangent to $AC$ at $A$ (as $\angle ABX = \angle BAX = \angle CAX$), so $X$ is the $B$-HM point and the conclusion is well-known.

Let $M$ be the midpoint of $AC$ and $BB'$. The $AXCB'$ is cyclic as $$\angle AXB'=\angle XAB+\angle XBA=\angle BAC=\angle B'CA$$Notice we have that $HAB'C$ is also cyclic. As $X$ lies on the $B$-median this uniquely defines $X$ as the $B$-Humpty point, implying the result.

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