This is my latest version.
Lemma #1 Let HRLI and ARLJ be two intersecting circles with HRA and JLI being straight lines. Then, AJ // HI. Proof is skipped
Lemma #2 Let A, R, L be inscribed in a circle with AL as the diameter and O as the center (i.e. $\angle ARL = 90^0$). $\triangle RLJ$ is similarly defined with $\angle RLJ = 90^0$. Then, ARLJ is cyclic. The proof is skipped.
As mentioned earlier, since XZ = 2BM = 2MC = ZY, we only need to show $TZY =90^0$.
ABD, ACE, and AM are respectively extended to X, Y and Z such that AB = BX, AC = CY, and AM = MZ.
We have three intersecting circles and they are (1) The red circle M (diameter = AMZ); (2) The blue circle B (diameter = ABX); and (3) The cyan circle C (diameter = ACY). They share the same common chord AH. With respect to AH, CMB being the line of centers, will bisect AH perpendicularly at R. That is, CRMB is a straight line.
By angles in semi-circle, $\angle YHA = \angle AHZ = \angle AHX = 90^0$. This in turn means XZHY is a straight line.
The orange circle O (diameter = AOT) cuts circle M at A and K. That is, AK is their common chord with GMOU being the corresponding line of centers (where G and U are points as shown). By angle in semi-circle, $\angle AKZ = \angle AKT = 90^0$. This, in turn, means TZK is a straight line.
(Please skip the next two paragraphs.)
Draw the green circle (centered at B, radius = BZ). Produce ZB to cut the same circle at J. That is, Z(B)J is the diameter. Note that $\angle ZKJ = 90^0$ and AKJ is a straight line. Then, BK = BZ. The above, together with MK = MZ, implies MKBZ is a kite with $\angle MNK = 90^0$.
Finally, KNMG, GARM, MRHU, and UMNZ are rectangles in turn. The required result follows.
Produce ZB to cut AK produced at J. Draw JI perpendicular to XZ cutting X(ZUH)Y at I. Clearly, ZKJI is cyclic. Produce C(RMN)B to cut IJ at L. Since L is the 4th vertex of the potential rectangle IHRL, $\angle BLJ = \angle BLI = 90^0$. In addition, IHRL is cyclic.
Form the circle passing through A, R, L with diameter = AL. Also, form the circle passing through R, L, J with diameter = RJ. (Some conditions need to be added before Lemma 2 can be applied, ARLJ is cyclic. TO BE FIXED). By lemma 1, AJ // HI.
The required result follows.