18.3 In a convex quadrilateral $ABCD$, it is known that $\angle DAB = \angle BCD = 90^{\circ}$, and $\angle CDA : \angle ABC = 2 : 1$, $AD : CB = 1 : \sqrt{3}$. Then the degree measure of $\angle BDA$ is
(A) $30^{\circ}$.
(B) $45^{\circ}$.
(C) $60^{\circ}$.
(D) cannot be determined.
(1st "Hope Cup" National Mathematics Invitational Competition, 1990)
[Solution] As shown in the figure, from $\angle C D A + \angle A B C = 180^{\circ}$ and the given ratio $\angle C D A : \angle A B C = 2 : 1$, we have
$$
\angle C D A = 120^{\circ}, \angle A B C = 60^{\circ}.
$$

Extend $C D$ to intersect $B A$ at $E$, and connect $B D$.
Then, since $\angle C = 90^{\circ}$, we get $\angle E = 30^{\circ}$.
Let $A D = x$, then $B C = \sqrt{3} x$, and $D E = 2 x$.
Also, $B E = 2 B C = 2 \sqrt{3} x$, so $C E = 3 x$.
Thus, $C D = x$, and at this point $\angle C D B = 60^{\circ}$, then $\angle B D A = 60^{\circ}$. Therefore, the answer is $(C)$.