6. In an acute triangle $ABC$, the altitudes from $AC$ and $AB$ are $BE$ and $CF$ respectively, with $E$ and $F$ being the feet of the perpendiculars, and $BE$ and $CF$ intersect at $H$. If $\angle BAC=60^{\circ}, BC=2$, then the length of $AK$ is ( ).
(A) $\frac{3}{2}$.
(B) $\frac{\pi}{3}-\sqrt{3}$
(C) $\frac{\sqrt{2}}{2}$
(D) 1
6. (B).

In the right-angled $\triangle A B E$, $A E$ $=A B \cdot \cos \angle B A C$. Similarly, $A F=A C \cdot \cos \angle B A C$.
In $\triangle A E F$,
$$
\begin{array}{l} 
E F^{2}= A E^{2}+A F^{2} \\
-2 A E \cdot A F \cdot \cos \angle B A C \\
= \cos ^{2} \angle B A C\left(A B^{2}+A C^{2}\right. \\
-2 A B \cdot A C \cdot \cos \angle B A C) \\
= B C^{2} \cdot \cos ^{2} \angle B A C .
\end{array}
$$

Given $B C=2, \angle B A C=60^{\circ}$, then $E F=2 \cos 60^{\circ}=1$.
Since $A, F, H, E$ are concyclic, $A H$ is the diameter of the circle. Let it be $d$, then by the Law of Sines $E F=d \cdot \sin \angle B A C$, i.e., $d=\frac{E F}{\sin \angle B A C}=\frac{2}{\sqrt{3}} = \frac{2}{3} \sqrt{3}$.