Japan Mathematical Olympiad

For distinct three points $P$, $Q$, $R$, the description $\angle QPR=\theta$ means that line $PQ$ rotated around $P$ by angle $\theta$ counterclockwise coincides with line $PR$. Here $180^{\circ }$ difference is ignored. By the inscribed angle theorem, we have $\angle XBD=\angle XEC$, $\angle XDB=\angle XCE$. We also have $BD=BM=CM=CE$ hence triangle $BDX$ and triangle $ECX$ are congruent, therefore $BX=EX$. Let $N$ be the midpoint of arc $BC$ of triangle $ABC$'s circumcircle which includes $A$. We have $\angle BXE=\angle BAE=\angle BAC=\angle BNC$, and triangle $BEX$ is an isosceles triangle with apex $X$, triangle $BCN$ is an isosceles triangle with apex $N$. Also both $\angle BXE$ and $\angle BAC$ are acute angles, hence those two triangles are similar including orientation. Therefore we obtain $BX:BE=BN:BC$ and $\angle NBX=\angle EBX+\angle NBE=\angle CBN+\angle NBE=\angle CBE$, which shows triangle $BNX$ and triangle $BCE$ are similar. Also we have $$\angle XAB=\angle XEB=\angle NCB=\angle NAB$$ thus the points $A$, $N$, $X$ are collinear. Furthermore, $\angle BNO=\angle BNM=\angle BCM$ holds and triangle $BNO$ is an isosceles triangle with apex $O$, triangle $BCM$ is an isosceles triangle with apex $M$ hence those two are similar. Therefore by $CE=CM=BM$ we obtain $$NX=CE\cdot \frac{BN}{BC}=CM\cdot \frac{BO}{BM}=NO.$$ Let $T$ be the point symmetric to $N$ with respect to line $OX$, then above shows $TO=NO=NX=TX$ hence quadrilateral $NOTX$ is a rhombus. Therefore we have $\angle OTX=\angle XNO=\angle OAX$, thus $T$ is on the circumcircle of triangle $AOX$. Also by $NO=OT$, $T$ is on the circumcircle of triangle $ABC$. Since we have $\angle MBD=\angle MCE$ and $BD=BM=CM=CE$, triangle $BDM$ and triangle $CEM$ are congruent, implying $DM=EM$. Also we have $\angle ADM=\angle CEM$ thus $E$ is on the circumcircle of triangle $ADM$. Furthermore, line $NX$ and line $OT$ are parallel and line $NX$ and line $AM$ are orthogonal, hence line $OT$ and line $AM$ are also orthogonal. Thus by $AO=MO$, we obtain $AT=MT$. Therefore, $A$ and $M$ are symmetric with respect to line $OT$, hence we have $\angle OAT=\angle TMO=\angle OTM$ and thus line $MT$ is tangent to the circumcircle of triangle $AOT$. Since we have $$\angle ATM=\angle ABM=\angle ADM+\angle DMB=2\angle ADM$$

and two points $D$ and $T$ are on the same side with respect to line $AM$, $T$ is the circumcenter of triangle $ADM$. Therefore, we have $DT=ET$, thus by $DM=EM$ we have proved line $MT$ is the perpendicular bisector of segment $DE$.