4. In $\triangle A B C$, points $D$ and $E$ are on sides $A B$ and $A C$ respectively, and $B E$ intersects $C D$ at point $O$. Then the size relationship between $\frac{S_{\triangle A D E}}{S_{\triangle A B C}}$ and $\frac{S_{\triangle D O E}}{S_{\triangle B O C}}$ is $\qquad$ (fill in "equal" or "not equal").
4. Equality.

As shown in Figure 6, let
$$
\begin{array}{l}
S_{\triangle A D E}=a, S_{\triangle D O E}=b, S_{\triangle B O C}=c, \\
S_{\triangle A O D}=x, S_{\triangle C O E}=y . \\
\text { By } \frac{S_{\triangle B O D}}{S_{\triangle D O E}}=\frac{B O}{O E}=\frac{S_{\triangle B O C}}{S_{\triangle C O E}}, \text { we get } \\
\frac{x}{b}=\frac{c}{y} \Rightarrow x y=b c . \\
\text { By } \frac{S_{\triangle B D E}}{S_{\triangle A D E}}=\frac{B D}{A D}=\frac{S_{\triangle B C D}}{S_{\triangle M C D}}, \text { we get } \\
\frac{b+x}{a}=\frac{c+x}{a+b+y} \\
\Rightarrow b(x+y)+x y=a c-a b-b^{2} .
\end{array}
$$

From equations (1) and (2), we solve for
$$
x+y=\frac{a c-a b-b^{2}-b c}{b} \text {. }
$$

Therefore, \( S_{\triangle A B C}=a+b+c+(x+y)=\frac{a c}{b} \).
$$
\text { Hence } \frac{S_{\triangle A D E}}{S_{\triangle A B C}}=a \cdot \frac{b}{a c}=\frac{b}{c}=\frac{S_{\triangle D O E}}{S_{\triangle B O C}} \text {. }
$$