$17 \cdot 84$ In $\triangle A B C$, $A B=2 \sqrt{3}, \quad A C=2$, and the altitude from $A$ to $B C$ is $\sqrt{3}$. Then the length of $B C$ is
(A) 4.
(B) 2.
(C) 4 or 2.
(D) None of the above.
(Chinese Zhejiang Province Junior High School Mathematics Competition, 1993)
[Solution] From the given and Fig. (2), we have
$$
\begin{array}{l}
B H=\sqrt{A B^{2}-\overline{A H^{2}}}=\sqrt{12-3}=3, \\
\text { Also, } \quad C H=\sqrt{A C^{2}-A H^{2}}=\sqrt{4-3}=1 \text {. } \\
\end{array}
$$

From Fig. (1) and (2), we know $B C=4$ or 2. Therefore, the answer is $(C)$.