This is the 3rd version.

ABD, ACE, and AM are respectively extended to X, Y, and Z such that AB = BX, AC = CY, and AM = MZ.

By midpoint theorem, XZ = 2BM = 2MC = ZY and XZY is a straight line parallel to BMC.

Since, by midpoint theorem, $OB = \dfrac 12 TX$ and $OC = \dfrac 12 TX$, we need to prove that TX = TY.
Adding the fact stated above, we only need to show $\angle TZY = ….= 90^0$.

The red circle (centered at M, radius = AM, diameter = AMZ) will cut XZY at G such that $\angle AGZ = 90^0$.

The green circle (centered at O, radius = AO, diameter = AOT) will cut TY at K such that $\angle AKT = 90^0$.

The purple circle (centered at C, radius = AC, diameter = ACY) will cut TY at K also such that $\angle AKY = 90^0$.

The blue circle (centered at B, radius = AB, diameter = ABX) will cut XZY at G such that $\angle AGZ = 90^0$.

Note that the grey circle can also be formed (because $\angle MET = \angle MDT = 90^0$) passing through M, E, T, D with MT as diameter. Then, when TZ is produced, it will cut MC at H such that $\angle MHT = 90^0$ (because $\angle MHT $ subtends the diameter MT).

Note that AG is the chord common to circles C, M, B. CHMB is then a straight line (the line of centers) and it will cut the common chord AG perpendicularly at G’.

Result follows because Z is the fourth vertex of the rectangle HG’GZ.