How to place points $D$ on side $AB$ and $E$ on side $AC$ of triangle $ABC$ so that the areas of triangles $ADE$, $DEB$, and $EBC$ are equal?
The heights from vertex $E$ in triangles $A D E$ and $B D E$ are common, so their areas can only be equal if their bases are also equal. Therefore, point $D$ must be chosen such that $A D = D B$, regardless of the position of $E$.

![](https://cdn.mathpix.com/cropped/2024_05_02_ac0ce98d6fa0610fab83g-1.jpg?height=344&width=542&top_left_y=300&top_left_x=781)

The combined area of the three parts gives the total area $T$ of triangle $A B C$. Since the three parts are equal, one part is exactly $(1/3)T$. The triangle $B E C$ shares a side with triangle $A B C$, and thus the ratio of their areas is equal to the ratio of the heights corresponding to the common side $B C$. From this, it follows that the ratio of the corresponding sides is $C E : C A = 1 : 3$.

Therefore, point $D$ bisects $A B$, and point $E$ is the point that divides $A C$ into a ratio of $1:2$, closer to $C$.