2. As shown in Figure 1, given that $E$ and $F$ are the midpoints of sides $AD$ and $CD$ of a square $ABCD$ with side length 6, $AF$ and $BE$ intersect at point $P$. Then the length of $PC$ is ( ).
(A) 6
(B) 6.5
(C) 5
(D) 5.5
2. A.

As shown in Figure 2, extend $A F$ and $B C$ to intersect at point $G$. It is easy to prove
Rt $\triangle A B E$
$\cong \operatorname{Rt} \triangle D A F$
$\cong \mathrm{Rt} \triangle C G F$.

Thus, $B C=A D=C G, \angle A P B=90^{\circ}$. Therefore, $P C=\frac{1}{2} B G=B C=6$.