5. As shown in Figure 2, in equilateral $\triangle A B C$, $B D=2 D C$, $D E \perp B E, C E$ and $A D$ intersect at point $P$. Then ( ).
(A) $A P>A E>E P$
(B) $A E>A P>E P$
(C) $A P>E P>A E$
(D) $E P>A E>A P$
5.A.

From $B E=\frac{1}{2} B D=D C$, we have $\triangle B E C \cong \triangle C D A$.
Then $\angle A P E=\angle P A C+\angle A C P=\angle P C D+\angle A C P=60^{\circ}$.
Also, $\angle E A P A E>E P$.
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