3. In a convex quadrilateral $A B C D$, points $E$ and $F$ are the midpoints of sides $B C$ and $C D$ respectively. Segments $A E, A F$, and $E F$ divide the quadrilateral into 4 triangles, the areas of which are consecutive natural numbers. What is the maximum possible value of the area of triangle $A B D$?
3. Let the areas of the triangles be $n, n+1, n+2$, $n+3$. Then the area of the quadrilateral $A B C D$ is $4 n+6$. It is easy to see that the area of triangle $B C D$ is four times the area of triangle $E C F$, so this area is at least $4 n$. Therefore,

$$
S_{A B D}=S_{A B C D}-S_{B C D} \leqslant(4 n+6)-4 n=6 .
$$

Equality is achieved if triangle $E C F$ has the smallest area

![](https://cdn.mathpix.com/cropped/2024_05_21_6455f48598646f41a890g-307.jpg?height=177&width=333&top_left_y=1048&top_left_x=911)

Fig. 151 of the four triangles mentioned.

It remains to prove that the value 6 is possible. An example is an isosceles trapezoid with bases $A D=6, B C=4$ and height 2 (the numbers in Fig. 151 denote the areas of the triangles).

Comment. It is easy to see that the area of triangle $A B D$ is always either 2 or 6.