Problem 1494

1. Vector Representation and Setup:
- Choose the origin $I$ at the intersection of diagonals $AC$ and $BD$.
- Let $\frac{IC}{IA} = a > 0$ and $\frac{ID}{IB} = b > 0$.
- This implies $C = -aA$ and $D = -bB$.

2. Midpoints Calculation:
- The midpoint $M$ of $AB$ is given by:
$$M = \frac{A + B}{2}$$
- The midpoint $N$ of $BC$ is given by:
$$N = \frac{B + C}{2} = \frac{B - aA}{2}$$

3. Using Menelaus' Theorem:
- Apply Menelaus' theorem to the transversal $\overline{AEN}$ in triangle $BCI$:
$$\frac{AE}{EN} \cdot \frac{NB}{BC} \cdot \frac{CI}{IA} = 1$$
- Similarly, apply Menelaus' theorem to the transversal $\overline{DFM}$ in triangle $ABI$:
$$\frac{DF}{FM} \cdot \frac{MA}{AB} \cdot \frac{BI}{ID} = 1$$

4. **Finding $E$ and $F$:**
- From the given $BE = \frac{1}{3} BD$, we have:
$$E = \frac{1}{a+2}B$$
- From the given $AF = \frac{1}{3} AC$, we have:
$$F = \frac{b}{2b+1}A$$

5. Equating the Conditions:
- Given $3 \cdot BE = BD$ and $3 \cdot AF = AC$, we get:
$$E = \frac{2-b}{3}B \quad \text{and} \quad F = \frac{2-a}{3}A$$
- Equate the expressions for $E$ and $F$:
$$\frac{1}{a+2} = \frac{2-b}{3} \quad \text{and} \quad \frac{b}{2b+1} = \frac{2-a}{3}$$

6. Solving the Equations:
- Solve the first equation:
$$3 = (2 - b)(a + 2) \implies 3 = 2a + 4 - ab - 2b \implies ab = 2(a - b) + 1$$
- Solve the second equation:
$$3b = (2 - a)(2b + 1) \implies 3b = 4b + 2 - 2ab - a \implies 2ab = b - a + 2$$

7. Simplifying the Equations:
- Combine the equations:
$$ab = 2(a - b) + 1 \quad \text{and} \quad 2ab = b - a + 2$$
- Solving these simultaneously, we get:
$$a = b = 1$$

8. Conclusion:
- Since $a = b = 1$, it follows that $IC = IA$ and $ID = IB$.
- Therefore, the quadrilateral $ABCD$ is a parallelogram.

$\blacksquare$