18.67 In $\triangle A B C$, $C A=C B$, and a square $B C D E$ is constructed outward from side $B C$ of $\triangle A B C$. If the degree measure of $\angle D A B$ is $\alpha$, then
(A) $\alpha$ depends on $\triangle A B C$.
(B) $\alpha$ is independent of the triangle.
(C) $\alpha<\angle C A B$.
(D) $\alpha=\angle C A D$.
(E) $45^{\circ}<\alpha<90^{\circ}$.
［Solution］Let $\angle C A D=\beta$, by the property of the sum of internal angles of a triangle, we have
$$
\left[180^{\circ}-2(\alpha+\beta)\right]+90^{\circ}+2 \beta=180^{\circ} \text {, }
$$

which gives $\alpha=45^{\circ}$. That is, $\alpha$ is independent of the shape of the isosceles triangle. Therefore, the answer is $(B)$.