5. As shown in Figure 2, in square $ABCD$, $E$ and $F$ are points on sides $AB$ and $BC$ respectively. $DE$ intersects $AC$ at point $M$, and $AF$ intersects $BD$ at point $N$. If $AF$ bisects $\angle BAC$, $DE \perp AF$, and let $x=\frac{BE}{OM}$, $y=\frac{BN}{ON}$, $z=\frac{CF}{BF}$, then ( ).
(A) $x>y>z$
(B) $x=y=z$
(C) $x=y>z$
(D) $x>y=z$
5. D.

By the Angle Bisector Theorem, we have
$$
\frac{B N}{O N}=\frac{A B}{A O}=\sqrt{2}=\frac{A C}{A B}=\frac{C F}{B F},
$$

which means $y=z=\sqrt{2}$.
Since the angle bisector of $\triangle A M E$ coincides with its altitude, $\triangle A M E$ is an isosceles triangle, i.e., $A M=A E$.
As shown in Figure 4, draw $O P / / A B$,

intersecting $D E$ at point $P$. Then $O P$ is
the midline of $\triangle D B E$.
Thus, $\triangle O M P \backsim \triangle A M E$
$\Rightarrow x=\frac{B E}{O M}=\frac{B E}{O P}=2$.
In summary, $x>y=z$.