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I have trouble solving this problem:

If $E$ is the midpoint of the segment $BC$, $L_1 \parallel L_2$, $AD=4cm,\ DE=5cm\ ,EA=7cm$, find the area of the quadrilateral $ABCD$.

Triangle inside quadrilateral ABCD

The answer is $8\sqrt{6}$.

I calculated the triangle's area using the Heron's formula, but that's it, if I use the theorem of a triangle inside a parallelogram using the area that I calculated, I get the correct answer, but since the shape is not one (from what I can see), that could be just a coincidence. I haven't used the fact that $E$ is the midpoint of the $BC$ segment, so I think that info makes the shape a parallelogram, and that's what makes the theorem work, but I don't know why. Thanks.

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  • $\begingroup$ Draw a line parallel to AD and passing through $E$. Suppose that that line intersects L1 and L2 at C' and B' respectively. The parallelogram AB'C'D has the same area as the trapezoid because the triangle EB'B and CEC' are equal. The triangle AED has half the area of the parallellogram AB'C'D. $\endgroup$ – Bettybel Jul 5 '17 at 20:04
  • $\begingroup$ @Bettybel Now I get it! Would you write your comment as an answer? I had less trouble understanding yours. $\endgroup$ – Nick Cassol Jul 5 '17 at 20:22
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Draw a line parallel to $AD$ and passing through $E$. Suppose that that line intersects $L1$ and $L2$ at $C'$ and $B'$, respectively.

The triangle $AED$ has half the area of the parallellogram $AB'C'D$.

The parallelogram $AB'C'D$ has the same area as the trapezoid because the triangle $BEB'$ and $CEC'$ are equal (ALA).

Therefore the triangle $AED$ has half the area of the trapezoid $ABCD$ too.

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Call the perpendicular distance between the lines $k$. Then the area of $\triangle CDE$ is $\frac{1}{2} CD\frac{k}{2}=\frac{CDk}{4}.$ The area of $\triangle AEB$ is $\frac{1}{2} AB\frac{k}{2} = \frac{ABk}{4}.$ The area of the trapezoid $ABCD$ is $k\frac{CD+AB}{2}$.

Comparing these areas shows that $\triangle ADE$ is half the area of the trapezoid.

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