3. Given that the diagonals of $\square A B C D$ intersect at point $O$, $\angle A D C=40^{\circ}$, and $E$ is a point on side $B C$ such that $A D-A B=$ $2 B E$. Then the degree measure of $\angle B E O$ is ( ).
(A) $140^{\circ}$
(B) $150^{\circ}$
(C) $160^{\circ}$
(D) $165^{\circ}$
3. C.

As shown in Figure 3, take the midpoint $F$ of $BC$, and connect $OF$.
$$
\begin{array}{l}
\text { Then } BF=\frac{1}{2} BC, \\
OF=\frac{1}{2} AB, \\
OF \parallel AB .
\end{array}
$$

Therefore, $\angle OFC=\angle ABC=\angle ADC=40^{\circ}$.
Since $AD-AB=2BE$, we have,
$$
\begin{array}{l}
OF=\frac{1}{2} AB=\frac{1}{2} AD-BE \\
=BF-BE=EF .
\end{array}
$$

Thus, $\angle OEF=\angle EOF=\frac{1}{2} \angle OFC=20^{\circ}$.
Hence, $\angle BEO=180^{\circ}-\angle OEF=160^{\circ}$.