5. As shown in Figure 2, points $E$ and $F$ are on the sides $BC$ and $AB$ of rectangle $ABCD$, respectively, with $BF=3$, $BE=4$, $CE=3$, and $AE$ intersects $CF$ at point $P$. Given that $\angle APC = \angle AEB + \angle CFB$, the area of rectangle $ABCD$ is ( ).
(A) 70
(B) 63
(C) 77
(D) 65.8
5. A.

As shown in Figure 6, draw $E Q / / C F$, and draw $F Q / / B C$. Connect $A Q$ and $B Q$.
Notice that
$\angle A P C$
$=\angle A E B +$
$\angle C F B$,
$\angle A P C = \angle E P F$,
$\angle A B C = 90^{\circ}$,

then $\angle E P F = 135^{\circ}$,
$\angle A P F = 45^{\circ}$.

Thus, $\angle Q E A = \angle A P F = 45^{\circ}$,
$\angle B C F = \angle B E Q$,
$\angle B F Q = \angle A B C = 90^{\circ}$.
Since quadrilateral $C E Q F$ is a parallelogram, we have
$F Q = C E = 3 = B F$.
Therefore, $\angle A B Q = 45^{\circ} = \angle Q E A$.
Hence, points $A$, $E$, $B$, and $Q$ are concyclic.
Thus, $\angle B A Q = \angle B E Q = \angle B C F$.
Therefore, Rt $\triangle F A Q \cong \mathrm{Rt} \triangle B C F$.
This gives $A F = B C = B E + C E = 7$.
Thus, $A B = A F + B F = 10$.
Therefore, the area of rectangle $A B C D$ is
$A B \cdot B C = 10 \times 7 = 70$.