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Here is a problem from a sample paper:enter image description here

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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marked as duplicate by Parth Kohli, Claude Leibovici, egreg, TZakrevskiy, M Turgeon Mar 28 at 12:01

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1 Answer 1

up vote 2 down vote accepted

Hint: The triangle ABE and the quadrilateral ABCD share the area of the triangle ABC. It suffices to show the remaining areas are equal, which are the areas of the triangle ACD and that of the triangle ACE. How may you show this? Find another area that is equal to both that of ACD and ACE, for example.

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