2. As shown in Figure 1, the diagonals of
rectangle $A B C D$ intersect at point $O, A E$ bisects
$\angle B A D$ and intersects $B C$ at
point $E$. If $\angle C A E=$
$15^{\circ}$, then $\angle B O E=$ ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$
2. D.

From the given conditions, we know $\angle B A E=45^{\circ}$. Thus,
$$
\angle B A O=60^{\circ}, \angle B E A=45^{\circ} \text {. }
$$

Therefore, $\triangle A B O$ is an equilateral triangle, and $\triangle A B E$ is an isosceles right triangle.
Hence, $B E=A B=B O, \angle A B O=60^{\circ}$.
Thus, $\angle O B E=30^{\circ}$.
So, $\angle B O E=\angle O E B$
$$
=\frac{1}{2}\left(180^{\circ}-\angle O B E\right)=75^{\circ} .
$$