25. As shown in the figure, in $\triangle A B C$, $\angle C A B-\angle B=90^{\circ}, D$ is on the extension of $B C$, $C E$ bisects $\angle A C D$ and intersects the extension of $B A$ at $E$, then the degree of $\angle E$ is ( ).
(A) $30^{\circ}$
(B) $35^{\circ}$
(C) $40^{\circ}$
(D) $45^{\circ}$
Answer: D.
Solution: As shown in the figure,
it is obvious that
$$
\angle 1=\angle 2 \text {. }
$$

By the exterior angle theorem of a triangle, we get
$$
\begin{array}{c}
\angle 1=\angle B+\angle E, \\
\angle C A B=\angle E+\angle 2 .
\end{array}
$$

Since
$$
\angle 1=\angle 2 \text {, }
$$

and
$$
\angle C A B-\angle B=90^{\circ} \text {, }
$$

thus
$$
\begin{array}{c}
\angle C A B=\angle B+2 \angle E, \\
\angle E=\frac{1}{2}(\angle C A B-\angle B)=45^{\circ} .
\end{array}
$$

Therefore, the answer is D.