4. Given that $a, b, c$ are the three sides of a right triangle $\triangle A B C$, where $c$ is the hypotenuse. For $0<m<a+b-c$. If the sides $a, b, c$ are each shortened by the same length $m$, then the triangle formed by the segments $a-m, b-$ $m, c-m$ is ( ).
(A) acute triangle
(B) right triangle
(C) obtuse triangle
(D) acute, right, or obtuse triangle is possible
4. C
$$
\begin{array}{l}
\because(a-m)+(b-m)-(c-m) \\
\quad=a+b-c-m>0(\because m(c-m) .
\end{array}
$$

Similarly, $(a-m)+(c-m)>(\dot{b}-m)$,
$$
(b-m)+(c-m)>(a-m) .
$$
$\therefore$ A triangle with sides $a \cdots m, \dot{b}-m, c-m$ exists. Let such a triangle be $\triangle A^{\prime} B^{\prime} C^{\prime}$, with $c-m$ being the largest side of $\triangle A^{\prime} B^{\prime} C^{\prime}$.
$$
\begin{array}{l}
\therefore \cos C^{\prime}=\frac{(a-m)^{2}+(b-m)^{2}-(c-m)^{2}}{2(a-m)(b-m)} \\
\quad=\frac{a^{2}+b^{2}-c^{2}-2(a+b-c) m+m^{2}}{2(a-m)(b-m)} \\
\because a^{2}+b^{2}=c^{2}, a-m>0, b-m>0, \\
\therefore \cos C^{\prime}<0, \text{ i.e., } \angle C^{\prime} \text{ is an obtuse angle. }
\end{array}
$$
$\therefore \cos C^{\prime}<0$, i.e., $\angle C^{\prime}$ is an obtuse angle.
Thus, $\triangle A^{\prime} B^{\prime} C^{\prime}$ is an obtuse triangle.