$19 \cdot 51$ As shown, $\angle A O D=90^{\circ}$, and $O A=O B=B C=C D$. Then which of the following conclusions is correct?
(A) $\triangle A O B \backsim \triangle C O A$.
(B) $\triangle O C A \backsim \triangle O D A$.
(C) $\triangle B C A \backsim \triangle B A D$.
(D) None of the above conclusions is correct.
[Solution] Consider the right figure, and let $O A=1$, then $A B=\sqrt{2}, B D=2$. From $\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$,

we know $\frac{B C}{A B}=\frac{A B}{B D}$, that is, $\triangle B C A \backsim \triangle B A D$.
Therefore, the answer is (C).