$7 \cdot 34$ As shown, in $\triangle A B C$, $A B=A C, \angle A=36^{\circ}$. To make $\triangle A B C \backsim \triangle B C D$, the condition that needs to be added is
(A) $B C=C D$.
(B) $\frac{B D}{A C}=\frac{C D}{B C}$.
(C) $A D=D C$.
(D) $\frac{B C}{\sin A}=\sin 18^{\circ}$.
(Shanxi Province, China Junior High School Mathematics Competition, 1990)
[Solution] When $\angle B D C=\angle C=72^{\circ}$, i.e., $B D=B C$, if $\frac{B D}{A C}=\frac{C D}{B C}$, then $\triangle A B C \sim \triangle B C D$.
$\because \angle A=36^{\circ}$, then $\angle C=72^{\circ}, \angle B=72^{\circ}$.
If $\frac{B D}{A C}=\frac{C D}{B C}$, and by the Law of Sines, $\frac{\sin 72^{\circ}}{\sin \angle D B C}=\frac{\sin 72^{\circ}}{\sin 36^{\circ}}$,
$\because \angle D B C$ is an acute angle, $\therefore \angle D B C=36^{\circ}$.
Then $\angle B D C=\angle C=72^{\circ}$.
Therefore, the answer is $(B)$.