$21 \cdot 14$ As shown in the figure, an equilateral triangle $A B C$ is folded so that vertex $A$ falls on point $A^{\prime}$ on side $B C$. If $B A^{\prime}=1, A^{\prime} C=2$, then the length of the fold line $P Q$ is
(A) $\frac{8}{5}$.
(B) $\frac{7}{20} \sqrt{21}$.
(C) $\frac{1+\sqrt{5}}{2}$.
(D) $\frac{13}{8}$.
(E) $\sqrt{3}$.
(42nd American High School Mathematics Examination, 1991)
[Solution] Let $A P=x$, then $B P=3-x$; Let $A Q=y$, then $Q C=3-y$. - In $\triangle A^{\prime} B P$, since $A^{\prime} P=A P=x$, by the cosine rule we have
$$
x^{2}=(3-x)^{2}+1^{2}-2 \cdot 1 \cdot(3-x) \cos 60^{\circ} \text {. }
$$

Solving this, we get
$$
x=\frac{7}{5} \text {. }
$$

Similarly, in $\triangle A^{\prime} Q C$, we can solve for $A^{\prime} Q=y=\frac{7}{4}$.
Thus, in $\triangle A P Q$, by the cosine rule we have
$$
\begin{aligned}
P Q^{2} & =A P^{2}+A Q^{2}-2 A P \cdot A Q \cos 60^{\circ}=x^{2}+y^{2}-x y \\
& =\left(\frac{7}{5}\right)^{2}+\left(\frac{7}{4}\right)^{2}-\frac{7}{5} \cdot \frac{7}{4}=\frac{49 \cdot 21}{400} .
\end{aligned}
$$

Therefore,
$$
P Q=\sqrt{\frac{49 \cdot 21}{400}}=\frac{7}{20} \sqrt{21} \text {. }
$$

Hence, the answer is (B).