I will give a proof for the problem in the OP by also adding some bonus properties. The given geometric constellation can be embedded in the following one, isolated in a new statement below that uses the same notation from the OP for a simple parallel (against my will). The bonus properties come according to my principle that if two lines are perpendicular, and if the point "has a meaning", then a faithful proof should elucidate, or at least mention this. Full details are given, this makes the exposition longer. For a short exposition stop after the statement. (Which is longer than the proof, and fill in the details by a glance.)
Given:
- Let $\Delta ABC$ be a triangle with $AB<AC$. We denote by $H$, $O$ its orthocenter, respectively its circumcenter.
- The tangent in $A$ to the circumcircle $(O)=(ABC)$ intersects $BC$ in $P$, so $OA\perp AP$.
- Let $D,E,F$ be the feet of the altitudes in $\Delta ABC$ constructed from $A,B,C$, respectively. Let $M$ be the mid point of $BC$.
- We construct the Euler nine-points circle $(DEFM)=(9)$, centered in $9$, and would like also to single out some following further points on it. Let $X$ be the second point of intersection of $AM$ with $(9)$, here $X\ne M$. Let $Y\in(9)$ be the mid point of $AH$.
- Let $Q$ be the intersection of the parallel through $A$ to $BC$.
- Let $Z$ be the intersection of $EF$ with $BC$.
- Let $T$ be the intersection of $MY$ with $AP$.
- Let $\Xi$ be the intersection of $MH$ and $AZ$.
Then:
- (1) The lines $EF\| AP$ have a direction perpendicular to $AO\| M9YT$.
- (2) $Y$ is the orthocenter of $\Delta APM$ and $P,Y,X$ are colinear.
- (3) $H$ is the orthocenter of $\Delta AZM$, so in particular $ZH\|AYM$, both being perpendicular on $AXM$.
- (4) The triangles $\Delta YPD$ and $\Delta YQA$ are similar, in particular $P,Y(,X),Q$ are colinear, and their common line is perpendicular on $AM$ in $X$.
Proof:
(1) In the Euler circle we have $EF\perp MY$, since $MF=MB=MC=ME$ and $YF=YH=YA=YE$ (in a right triangle the median from the right angle is half hypotenuse), so $\Delta EMY=\Delta FMY$ so the chord $EF$ is perpendicular on the diameter $MY$ in $(9)$.
Then we have $EF\|AP$, since $EF$ is an "anti-parallel" to $BC$, using (blue) angles explicitly, $\widehat{PAB}=\frac 12\overset\frown{AB}=\hat C=\widehat{EFA}$.
And of course $PA\perp AO$ by construction.
(2) This is again simple. By (1) we have $MY\perp AP$, and of course $PBDMC$. So $Y$ is orthocenter in $\Delta APM$. The third height $AY$ intersects the side $AM$ in a point $X'$ such thet $\widehat YXM= 90^\circ$, so $X'=X$, since $YM$ is diameter in $(9)$.
(3) To be able to focus only on the few points involved, here is an other picture:
To have a simple proof, we reverse slightly the order of the construction of the points, with reference to the above picture. We construct first the points $A,B,C;D,E,F;H$ as above. Then we draw the circle $(AEFH)$ with diameter $AH$, and denote by $\Xi,\Upsilon$ the intersections of this circle with the lines $ZA$, $ZH$, respectively. Let $A\Upsilon$ intersect the line $PZBC$ in a point $M'$. (We want $M'=M$, so let us show it.) The using the power of the point $Z$ w.r.t. different circles, we have:
$$
\begin{aligned}
ZD\cdot ZM'
&=ZH\cdot Z\Upsilon &&\text{ powers of $Z$ w.r.t. the circle }(M'DHY)\\
&=ZE\cdot ZF &&\text{ powers of $Z$ w.r.t. the circle }(AFHE)\\
&=ZD\cdot ZM &&\text{ powers of $Z$ w.r.t. the circle }(9)\ .
\end{aligned}
$$
In a computer game, $M$ would inherit the property of $M'$, so $AH\Upsilon\perp AM$. Given that $AHD\perp AM$, the point $H$ is the orthocenter in $\Delta AZM$, and thus the third height $MH$ goes through the projection of $H$ on $AZ$, which is $\Xi$.
(4) We first observe that $APZQ$ is a parallelogram, $AQ\| PZ$ by construction, and $EF\|AP$. (Or use angle chasing to show that (two of) the "red angles" in $P,Q,Z$ have the measure $\hat B-\hat C$.)
The two triangles questioned about similarity have each a right angle in $A,D$, so it is enought to show a proportionality of two pairs of sides. We take the easy ones, and want to show (!):
$$
\begin{aligned}
\frac{AQ}{PD}
&\ \overset{(!)}= \
\frac{AY}{YD}\ . &&\text{ Equivalently:}
\\
\frac{PZ}{PD}
&\ \overset{(!)}= \
\frac{AY}{YD}\ . &&\text{ Equivalently, using derived proportions:}
\\
\frac{PZ}{PD-PZ}
&\ \overset{(!)}= \
\frac{AY}{YD-AY}\ . &&\text{ Equivalently:}
\\
\frac{PZ}{ZD}
&\ \overset{(!)}= \
\frac{YH}{HD}\ . &&\text{ True, because of }ZH\|PY\ .
\end{aligned}
$$
(Since $ZH$, $PY$ are perpendicular on $AM$ in $\Upsilon$, respectively $X$.)
The two angles in $Y$ of the two similar triangles are equal, given that $A,Y,D$ are colinear, we obtain $Q,Y,P$ colinear. And we already know $X$ is on the same line (with $P,Y$). The final touch is
$$
AXM\perp AYXQ\text{ in }X\ .
$$
$\square$
Best Answer
Let P be the midpoint of FB. Extend MP to cut TF at Q.
After such construction, MBQF is a kite with $\angle DFB = \angle DFQ = \angle FBQ$ and $\angle BQM = \angle FQM$. The latter implies QM is the external angle bisector of $\angle BQF$ of $\triangle TQB$.
Draw QR // AB cutting TB at R. Since $\angle RQM = \angle BPM = 90^0$, QR is then the internal angle bisector of $\angle TQB$.
Therefore, $\dfrac {TM}{BM} = \dfrac {TR}{RB}$.
But, $\dfrac {TR}{RB}= \dfrac {TQ}{QF}$; because of similar triangles.
Further, $\dfrac {TQ}{QF} = \dfrac {TB}{BD}$; because of similar triangles. .