42nd Balkan Mathematical Olympiad

Let $P$ be the midpoint of $AC$. Applying the formula for the length of the median on $△ABC$ and $△ADC$, and using the fact that $A{B}^2+B{C}^2=A{D}^2+C{D}^2$, we obtain $BP=DP$. $$\text{Claim.}\angle BXA=\angle PBD$$ 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 $△DUB$. Also observe that $A,U,B,X$ are concyclic ($\angle AUX=\angle ABX=90^{\circ }$), so we can infer that $$\angle PBD=90^{\circ }-\angle DUB=90^{\circ }-\angle XUB=90^{\circ }-\angle XAB=\angle BXA.$$ $\square$

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 $△BXA\sim △BYC$. 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 $△BAC\sim △BXY$ 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.