[Solution] Suppose $A E$ is folded onto $E C$, and $E F$ is the crease. Let $B E=x, \quad B C=t$. Then
$$
\begin{array}{l}
A E=E C=5-x, \quad \text { and } \\
D F=x, \quad A G=x, \\
G E=5-2 x .
\end{array}
$$

By the Pythagorean theorem, we have
$$
\begin{aligned}
t^{2} & =E F^{2}-E G^{2} \\
& =(\sqrt{6})^{2}-(5-2 x)^{2},
\end{aligned}
$$

Also, $t^{2}=C E^{2}-E B^{2}=(5-x)^{2}-x^{2}$,
Thus, combining the above two equations, we get $20 x^{2}-15 x+22=0$.
Solving for $x$, we get
$$
x=2 \text { or } 5 \frac{1}{2} \text {. }
$$

Since $5 \frac{1}{2}>5-5 \frac{1}{2}$, in $\mathrm{Rt} \triangle B E C$ it is not valid (the leg is greater than the hypotenuse).
By $x=2$, we get $B C=t=\sqrt{5}$.
Therefore, the answer is $(D)$.
$21 \cdot 13$ rectangle has one side of length 5, and the other side is less than 4. The rectangle is folded so that two opposite vertices coincide. If the length of the fold is $\sqrt{6}$, then the length of the other side is
(A) $\sqrt{2}$.
(B) $\sqrt{3}$.
(C) 2 .
(D) $\sqrt{5}$.
(E) $\sqrt{\frac{11}{2}}$.
(16th American High School Mathematics Examination, 1965)