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

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

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?

• The area of $BEF$ doesn't depend on the position of $F\in AD$ because it is equal $\frac{BE\cdot h}{2}$ where $h$ is the height of the parallelogram. – SMM Feb 15 at 18:03

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 $$ABCD$$.
Let $$FK$$ be an altitude of $$ABCD$$.
Thus, $$S_{ABCD}=BC\cdot FK=2BE\cdot FK=4\cdot\frac{BE\cdot FK}{2}=4\cdot126=504.$$