5. $a, b, c, d$ are positive numbers, $a>b>c>d$, and
$$
\begin{array}{l}
x=\sqrt{(a b+c d)(a-b)(c-d)}, \\
y=\sqrt{(a c+b d)(a-c)(b-d)}, \\
z=\sqrt{(a d+b c)(a-d)(b-c)} .
\end{array}
$$

Then the sides $x, y, z$ ( ).
(A) must form an acute triangle
(B) must form an obtuse triangle
(C) must form a right triangle
(D) may not form a triangle
5.C.

Let $a b+c d=\alpha, a c+b d=\beta, a d+b c=\gamma$. Then
$$
\begin{array}{l}
(a-b)(c-d)=\beta-\gamma, \\
(a-c)(b-d)=\alpha-\gamma, \\
(a-d)(b-c)=\alpha-\beta .
\end{array}
$$

The left side of the above three equations are all positive, so the right side is also positive.
Therefore, we have $\alpha>\beta>\gamma$.
And $x^{2}=\alpha(\beta-\gamma), y^{2}=\beta(\alpha-\gamma), z^{2}=\gamma(\alpha-\beta)$, then $x^{2}+z^{2}=y^{2}$.
Thus, $y>x, y>z$, and $x^{2}+z^{2}+2 x z>y^{2}$, which means $(x+z)^{2}>y^{2}$.
Therefore, $x+z>y$.
Hence, a right-angled triangle can be formed with $x, y, z$ as the sides, with $y$ being the hypotenuse.