Given: In $\Delta ABC$, a line $DE$ is drawn parallel to side $BC$, intersecting $AB$ at $D$ and $AC$ at $E$.

To Prove:

$$\frac{AD}{DB} = \frac{AE}{EC}$$

Construction: Join vertex $B$ to $E$ and vertex $C$ to $D$.

Proof:

Consider $\Delta BDE$ and $\Delta CDE$. These triangles lie on the same base $DE$ and are between the same parallels $DE$ and $BC$. Therefore, their areas are equal, i.e.,

$$ar(\Delta BDE) = ar(\Delta CDE)$$

Now, consider $\Delta ADE$ and $\Delta BDE$. These triangles share the same vertex $E$, meaning they have the same altitude with respect to bases $AD$ and $DB$. Thus, the ratio of their areas is equal to the ratio of their bases:

$$\frac{ar(\Delta ADE)}{ar(\Delta BDE)} = \frac{AD}{DB}$$

Similarly, for $\Delta ADE$ and $\Delta CDE$, which share the same vertex $D$, we have

$$\frac{ar(\Delta ADE)}{ar(\Delta CDE)} = \frac{AE}{EC}$$

Since $ar(\Delta BDE) = ar(\Delta CDE)$, the denominators in the area ratios are equal. Therefore, equating the two ratios, we get

$$\frac{AD}{DB} = \frac{AE}{EC}$$

∴ Hence proved.