4. In an acute triangle $\triangle A B C$, $M$ is the midpoint of side $A C$. The circle $\Gamma$ passing through points $B$ and $M$ intersects sides $A B$ and $B C$ at the second points $P$ and $Q$, respectively. Point $T$ satisfies that quadrilateral $B P T Q$ is a parallelogram. If $T$ lies on the circumcircle of $\triangle A B C$, find all possible values of $\frac{B T}{B M}$.
4. $\sqrt{2}$.

Let $P Q$ intersect $B T$ at point $S$, then $S$ is the midpoint of $P Q$ and $B T$.

As shown in Figure 3, let $B^{\prime}$ be a point on the extension of $B M$ such that $B M = M B^{\prime}$. Then quadrilateral $A B C B^{\prime}$ is a parallelogram.
$$
\begin{array}{l}
\text { By } \angle A B B^{\prime} = \angle P Q M, \\
\angle B B^{\prime} A = \angle B^{\prime} B C = \angle M P Q, \\
\text { then } \triangle A B B^{\prime} \sim \triangle M Q P \\
\Rightarrow A M, M S \text { are corresponding medians } \\
\Rightarrow \angle S M P = \angle M A B^{\prime} = \angle B C A = \angle B T A. \text { (1) } \\
\text { By } \angle A C T = \angle P B T, \\
\angle T A C = \angle T B C = \angle B T P,
\end{array}
$$

Then $\triangle T C A \sim \triangle P B T$
$\Rightarrow T M, P S$ are corresponding medians
$$
\Rightarrow \angle M T A = \angle S P T = \angle B Q P = \angle B M P. \text { (2) }
$$

We will discuss two cases.
(1) Point $S$ is not on $B M$, as shown in Figure 3.

Assume without loss of generality that points $S$ and $A$ are on the same side of line $B M$.
From equations (1) and (2), we get
$$
\begin{array}{l} 
\angle B M S = \angle B M P - \angle S M P \\
= \angle M T A - \angle B T A = \angle M T B \\
\Rightarrow \triangle B S M \sim \triangle B M T \\
\Rightarrow B M^{2} = B S \cdot B T = \frac{B T^{2}}{2} \\
\Rightarrow \frac{B T}{B M} = \sqrt{2} .
\end{array}
$$
(2) Point $S$ is on $B M$, as shown in Figure 4.

From equation (2), we get
$$
\begin{array}{l}
\angle B C A = \angle M T A = \angle B Q P = \angle B M P \\
\Rightarrow P Q \parallel A C, P M \parallel A T \\
\Rightarrow \frac{B S}{B M} = \frac{B P}{B A} = \frac{B M}{B T} \\
\Rightarrow B T^{2} = 2 B S \cdot B T = 2 B M^{2} \\
\Rightarrow \frac{B T}{B M} = \sqrt{2} .
\end{array}
$$