Let $P$ be the midpoint of $AC$. Applying the formula for the length of the median on $\triangle ABC$ and $\triangle ADC$, and using the fact that $AB^2 + BC^2 = AD^2 + CD^2$, we obtain $BP = DP$.
$$\text{Claim.} \quad \angle BXA = \angle PBD$$
*Proof.* We use directed angles.
Let $U$ be the projection of $A$ onto $DX$ and $N$ be the midpoint of $UD$.

Observe that $AU \parallel CD$ and $P, N$ are the midpoints of $AC, UD$, so $PN \parallel CD$. Since $DU \perp CD$, we get that $PU = PD = PB$, so $P$ is the circumcenter of $\triangle DUB$. Also observe that $A, U, B, X$ are concyclic ($\angle AUX = \angle ABX = 90^\circ$), so we can infer that
$$\angle PBD = 90^\circ - \angle DUB = 90^\circ - \angle XUB = 90^\circ - \angle XAB = \angle BXA.$$
$\Box$

By projecting $C$ onto $YD$, we can similarly prove that $\angle BYC = \angle PBD$, so $\angle BYC = \angle BXA$. We also have $\angle XBA = 90^\circ$ and $\angle YBC = 90^\circ$, meaning $\triangle BXA \sim \triangle BYC$. This is a spiral similarity centered at $B$ sending $AX$ to $CY$. It follows that $B$ is also the center of spiral similarity sending $AC$ to $XY$.

Since $XB \perp AB$ and $YB \perp CB$, the angle of rotation in the spiral similarity $\triangle BAC \sim \triangle BXY$ is $90^\circ$. This means we also have $AC \perp XY$, thus $T$ is the projection of $P$ onto $XY$, so $PT \perp TM$. Moreover, the spiral similarity sends $P$ to $M$, so $BP \perp BM$. We can similarly prove that $DP \perp DM$, so points $T, M, B, D$ lie on the circle with diameter $PM$ and we are done.