18.69 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 $x$, then
(A) $x$ depends on $\triangle A B C$.
(B) $x$ is independent of $\triangle A B C$.
(C) $x=\angle C A D$.
(D) $x$ cannot equal $\angle C A B$.
(E) $45^{\circ}<x<90^{\circ}$.
(7th American High School Mathematics Examination, 1956)
［Solution］From the given, we know
$$
\angle A C B=180^{\circ}-2 \angle C A B \text {. }
$$

And from $A C=C D$ we know
$$
\angle C A D=\frac{1}{2}\left[180^{\circ}-\left(\angle A C B+90^{\circ}\right)\right] \text {, }
$$

which means $\angle C A D=45^{\circ}-\frac{1}{2} \angle A C B=45^{\circ}-\frac{1}{2}\left(180^{\circ}-2 \angle C A B\right)$
$$
=\angle C A B-45^{\circ},
$$

Also, $\angle C A D=\angle C A B-x$, so $x=45^{\circ}$. Therefore, the answer is $(B)$.