Irish Mathematical Olympiad

We use the usual notation: $a = |BC|$, $b = |CA|$, $c = |AB|$ and $\beta = \angle CBA$.

Because $D$ is between $B$ and $C$, there exists a real number $\lambda$ satisfying $0 < \lambda < 1$ and $|BD| = \lambda a$. Then $|DC| = (1 - \lambda)a$ and because $AB \parallel DE$, the Intercept Theorem implies $|EC| = (1 - \lambda)b$. Hence, $|AE| = \lambda b$ and because $BG \parallel FE$, the Intercept Theorem implies that $|AF| = \lambda c$. This finally implies $|BF| = (1 - \lambda)c$.

Therefore, the area of the parallelogram $BDEF$ is equal to
$$|BD| \cdot |BF| \cdot \sin \beta = \lambda(1 - \lambda)ac\sin\beta.$$
Because $a, c, \beta$ are constant, it suffices to maximise
$$\lambda(1 - \lambda) = \frac{1}{4} - \left(\lambda - \frac{1}{2}\right)^2 \le \frac{1}{4}.$$
We now see clearly that the parallelogram $BDEF$ has maximal area when $\lambda = 1/2$, i.e. when $D, E, F$ are the mid-points of the the sides of $\triangle ABC$.