2. As shown in Figure 1, construct squares $A B E F$ and $A C G H$ outside the acute triangle $\triangle A B C$, and draw the altitude $A D$ from $A$ to the base $B C$, intersecting $F H$ at $K$. The relationship between $B C$ and $A K$ is ().
(A) $B C>2 A K$
(B) $B C=2 A K$
(C) $B C<2 A K$
(D) Cannot be determined
2. (B)

As shown in Figure 5, $\triangle A B C$ is rotated $90^{\circ}$ clockwise around point $A$ to get $\triangle A F M$. Given $A C = A H$, and $H, A, M$ are collinear, with $A$ being the midpoint of $MH$.
From $B C \perp A D$, we know that $F M \parallel A D$.
In $\triangle H M F$, since $A$ is the midpoint, and $A K \parallel M F$, $A K$ is the midline of the triangle, so $A K = \frac{1}{2} M F = \frac{1}{2} B C$, which means $B C = 2 A K$.