4. Given points $D$ and $E$ are on the sides $AB$ and $AC$ of $\triangle ABC$ respectively, and the extension of $DE$ intersects the extension of $BC$ at point $F$. Then $BD + BC$ $\qquad$ $CE + ED$ (fill in “$>$”, “<” or “$=$”).


4. $>$.

As shown in Figure 7, draw $D G / / B C$, and draw $C G / / A B$, intersecting at point $G$.

Then quadrilateral $B C G D$ is a parallelogram.
Thus, $D G = B C$,
$$
C G = B D \text{. }
$$

Let $C G$ intersect $E F$ at $H$.
In $\triangle D G H$,
$$
D G + G H > E D + E H ;
$$

In $\triangle E C H$,
$$
E H + C H > C E \text{. }
$$
(1) + (2) gives
$$
D G + (G H + C H) + E H > E D + C E + E H \text{, }
$$

which simplifies to $D G + C G > E D + C E$.
Therefore, $B C + B D > E D + C E$.