Incomplete solution. SEE EDIT and COMPLETE SOLUTION below
Your phrasing is really confusing. If you just draw from $P$ and $Q$ the lines parallel to $BC$ and $AB$ respectively, they will meet at a point $T$, forming a parallelogram. There are many configurations in which $T$ lies on $(ABC)$.
From your drawing and video, though, it appears that $T$ must lie on the perpendicular bisector of $AC$, which forces the arcs $\overset{\huge\frown}{AT}$ and $\overset{\huge\frown}{TC}$ to be equal. In that case $BT$ bisects $\angle ABC$, and therefore $BPTQ$ is in fact a rhombus, and the solution becomes unique.
In conclusion what we can show is the following.
Let $ABC$ be a triangle with median $BM$. Let $P$ and $Q$ be the intersections between the circle with diameter $BM$ and lines $AB$ and $BC$ respectively. Let also $T$ be the intersection between the perpendicular bisector of $AC$ and the circumcircle of $ABC$. Show that if $BPTQ$ is a parallelogram, then $BT/BM = \sqrt 2$.
Let $O$ be the center of the rhombus and $O_1\in BM$ be the center of $(BPQ)$.
- As stated earlier $\overset{\huge\frown}{AT} \cong \overset{\huge\frown}{TC}$ implies $\angle ABT \cong\angle TBC$.
- Then by ASA criterion $BTP \cong BTQ$. Hence $BPTQ$ is a rhombus.
- By SSS criterion we have $BPO_1 \cong BQO_1$. Therefore $\angle ABM \cong \angle MBC$.
- Comparing 1. and 2. we obtain that $B$, $T$, and $M$ are aligned.
- Since $BM$ is both median and internal bisector, $ABC$ is an isosceles triangle, and $BT$ perpendicularly bisects $AC$.
- From 5. we derive that $BT$ is a diameter of $(ABC)$ and hence $O$ is the center of $(ABC)$.
- WLOG let the smaller circle diameter be $\overline{BM} = 2$. If $\overline{BO} = x$. Then, similarity $BPO\sim MPO$ implies
$$\overline{PO} = \sqrt{x(2-x)}.$$
- Similarity $BPO\sim BAM$ and 7. give
$$\overline{AM} = 2\sqrt{\frac{2-x}{x}}.$$
- Now observe that $AO \cong BO$, so that Pythagorean Theorem on $AOM$ leads to the equation
$$(2-x)^2 + \frac{4(2-x)}{x} = x^2,$$
whose only valid solution is
$$x = \sqrt 2.$$
This yields
$$\boxed{\frac{\overline{BT}}{\overline{BM}} = \sqrt{2}}.$$
EDIT
Thanks to David K's comment, I noticed that all the configurations (not only the one in which $T$ bisects arc $AC$) lead to the same ratio $\sqrt 2$. I will try to work an a more complete solution. I leave for now the answer, though incomplete: it may be useful for others to work out the entire problem.
SOLUTION TO THE GENERAL CASE
Probably a bit of an overkill, but I could not find any better approach.
Suppose that point $T$ is determined as the fourth vertex of a parallelogram $BPTQ$. We want to show that if $T \in (ABC)$ then $\overline{BT}/\overline{BM} = \sqrt 2$.
Let us fix a coordinate system with the origin in $B$, and let, WLOG, $C(2,0)$. Suppose $A$ lies on the line with equation $y=mx$, and precisely let $A(2t,2mt).$
Then the perpendicular bisector of $AB$ has equation $y-mt=-\frac1m(x-t)$. From here, we can easily find the circumcenter of $(ABC)$, that is
$$O'\left(1,\frac{-1+t+m^2t}m\right),$$ and thus the equation of $(ABC)$
$$x^2+y^2-2x +2y\frac{1-t-m^2t}{m}=0.\tag{1}\label{1}$$
We also have $M(1+t,mt)$ and therefore
$$\overline{BM}^2 = (1+t)^2+ m^2t^2.\tag{2}\label{2}$$
Point $Q$ is the projection of $M$ on $BC$. So we get $Q(1+t,0)$.
Point $P$ is the projection of $M$ on $AB$. Intersection between line $y=mx$ and line $y-mt = -\frac{1}{m}(x-t-t)$ yields
$$P\left(\frac{m^2t+1+t}{m^2+1},m\frac{m^2t+1+t}{m^2+1}\right).$$
It is now straightforward to intersect lines $y=m\frac{m^2t+1+t}{m^2+1}$ and $y=m(x-1-t)$ to get
$$T\left(\frac{2m^2t+m^2+2+2t}{m^2+1},m\frac{m^2t+1+t}{m^2+1}\right).$$
Since $T$ lies on $(ABC)$, its coordinates satisfy \eqref{1}, which means, in particular, that
$$\overline{BT}^2 = x_T^2+y_T^2 = 2x_T-2y_T\frac{1-t-m^2t}{m}.\tag{3}\label{3}$$
Replacing in \eqref{3} the coordinates of $T$ we just found we derive
\begin{eqnarray}
\overline{BT}^2&=&2\cdot \frac{2m^2t+m^2+2+2t-(1+m^2t+t)(1-m^2t-t)}{m^2+1}\\
&=& 2 \cdot \frac{2m^2t+m^2+2+2t-[1-(m^2t+t)^2]}{m^2+1}=\\
&=&2 \cdot \frac{2t(m^2+1)+(m^2+1)+t^2(m^2+1)^2}{m^2+1}=\\
&=& 2 \cdot (2t+1+t^2+m^2t^2)=\\
&\stackrel{\eqref{2}}{=}& 2\overline{BM}^2.
\end{eqnarray}
Best Answer
Hint 1: $AE:EC=1:2$
Hint 2: What is $AC$?