What is the area of $ABCD$ parallelogram where $E$ is mid-point of BC and the area of $BEC$ is 126?

Question

$ABCD$ is a parallelogram. Point $E$ divides $BC$ into two equal lengths. If the area of $BEF$ is 126, what is the area of $ABCD$?

Source: Bangladesh Math Olympiad 2017 Junior Category.

I can not solve this problem. I am confused on the position of $'F'$. There is no information mentioned about it. Is there any lack of information in this question? Can anyone give me a hint?

Answer 1

Hint:

1. Consider the area of $\triangle BFC$, which shares the same height with $\triangle BEF$.
2. Draw an auxiliary line through $F$ parallel to $AB$ to see that the area found in step (1) is actually half of the area of the parallelogram $ABCD$.

Answer 2

Hint: CEF has the same area as BEF as they are both a half of the same parallelogram.

Answer 3

Let $FK$ be an altitude of $ABCD$.

Thus, $$S_{ABCD}=BC\cdot FK=2BE\cdot FK=4\cdot\frac{BE\cdot FK}{2}=4\cdot126=504.$$