# Prove that $\rm area(\triangle ABE) = area (ABCD)$ [duplicate]

Here is a problem from a sample paper:

I see that $\rm area(ACE) = area(DCE)$. I also see that $\rm AB$ is the common base for $\triangle \rm ABE$ and $\rm ABCD$. So this means that the height of $\rm ABCD$, i.e., $\rm AD$ is exactly the half of $\rm BE$. I can't seem to get around the rest, though.

It seems that $\rm ACED$ is a parallelogram, but that'd mean $\rm BC = CE \Rightarrow AD = DC$ which doesn't seem so looking at the figure. Could anyone please give me a complete proof using only the theorems:

1. Two quadrilaterals lying on the same base and between the same parallels have the same area.

2. Two triangles lying on the same base and between the same parallels have the same area.

I also know that the area of 1 is twice the area of 2.

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