Team Selection Test

As $\angle O_1{M}_{1}A=\angle O_1{M}_{2}A=90^{\circ }$, we have that $O_1{M}_{1}A{M}_2$ is a cyclic quadrilateral and the point $S$ is its circumcenter. Hence $S$ is the circumcenter of the triangle $A{M}_1{M}_2$.

Next we observe that the triangles $A{M}_1{M}_2$ and $M_4{M}_2{M}_1$ are congruent since $A{M}_1=M_4{M}_2=\frac{AD}{2}$ and $A{M}_2=M_4{M}_1=\frac{AB}{2}$. Therefore we obtain that the quadrilateral $S{M}_1{O}_3{M}_2$ is a rhombus and hence the line $M_1{M}_2$ is the perpendicular bisector of the line segment $[S{O}_3]$.

Similarly we can get that $M_1{M}_3$ is the perpendicular bisector of the line segment $[T{O}_1]$. As the points $M_1,M_2$ and $M_3$ are collinear, the result follows.