Nice problem although the angle bisector thing can be misleading. Clearly $O_3$ and $O_4$ are the nine point centers of $ABD$ and $ACD$. Hence, the reflections of $O_1$ and $O_2$ wrt $O_3$ and $O_4$ are the orthocenters of $ABD$ and $ACD$ respectively. Hence $O_3S$ and $O_4T$ are both parallel to the $A$ altitude. Clearly we have $O_3S \parallel O_4T$ and $M_2M_3 \perp O_3S$. Now we show that the straight line $M_2-M_1-M_3$ is the perpendicular bisector of $SO_3$ and $TO_4$. Note that $M_2S = \frac{BO_1}{2} = M_2O_3$ and $M_1S = \frac{DO_3}{2} = M_1O_3$ and therefore $M_1M_2$ is the perpendicular bisector of $SO_3$. Similarly we obtain that $M_1M_3$ is the perpendicular bisector of $TO_4$ and we are done.

Because $S$ is the circumcenter of $\Delta A{M_1}{M_2}$, so $S{M_2}{O_3}{M_1}$ is a rhombus, hence ${M_1}{M_2}$ is the perpendicular bisector of $S{O_3}$. Similarly ${M_1}{M_3}$ is the perpendicular bisector of $T{O_4}$, so $S{O_3}{O_4}T$ is an isosceles trapezoid.

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Nice little problem..but a bit too elementary for the TST. Note that $AM_2O_3M_1$ is cyclic with center $S$ and diameter $R_{ABD}$.So $SM_2=SM_2=O_3M_2=O_3M_2=\frac{R_{ABD}}{2}$ and consequently $SO_3 \perp M_2M_3$.Similarly $TO_4 \perp M_2M_3$ and $SO_3 \parallel TO_4$.Also see that $\triangle{SM_1T} \cong \triangle{O_3M_1O_4}$ since $SM_1=O_3M_1=\frac{R_{ABD}}{2}$,$TM_1=O_4M_1=\frac{R_{ADC}}{2}$ and $\angle{SM_1T}=\angle{O_3M_1O_4}=\angle{A}$.So $STO_4O_3$ is indeed an isosceles trapezoid,as desired.