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$ABCD$ is a parallelogram. E is a point lying on $AD$. $BE$ and $AC$ intersect at $F$. It is given that $AE:ED = 2:1$ and the area of $AEF = 8cm^2$. Find the area of $CDE$.

Really basic. Im sorry im helping my nephew.

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  • $\begingroup$ I thought it was basic but i really can't $\endgroup$ – Yau Kin Hoe Jan 26 '18 at 12:51
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Since, $\Delta AFE\sim\Delta CFB$, we obtain: $$\frac{AF}{FC}=\frac{AE}{BC}=\frac{AE}{AD}=\frac{2}{3}.$$

Thus, $$\frac{S_{\Delta AFE}}{S_{\Delta ACD}}=\frac{\frac{1}{2}AF\cdot AE\cdot\sin\measuredangle CAD}{\frac{1}{2}AC\cdot AD\cdot\sin\measuredangle CAD}=\frac{2\cdot2}{5\cdot3}=\frac{4}{15}.$$ Id est, $$S_{\Delta ECD}=\frac{1}{3}S_{\Delta ACD}=\frac{1}{3}\cdot\frac{15}{4}\cdot8=10.$$

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$\triangle AFE$ and $\triangle CFB$. Let the height of the parallelogram be $5y$, and the length be $3x$. Then $0.5\times 2x \times 2y=8$, so $2xy=8$. The shaded area is $0.5\times x\times 5y=2.5xy=10$.

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  • $\begingroup$ This is a non trigonometric approach, hope you like it. $\endgroup$ – QuIcKmAtHs Jan 26 '18 at 12:59
  • $\begingroup$ My solution we can write without trigonometry, of course. $\endgroup$ – Michael Rozenberg Jan 26 '18 at 18:02
  • $\begingroup$ Yes, that is why I upvoted your answer $\endgroup$ – QuIcKmAtHs Jan 27 '18 at 0:07
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Hint: try to think about how you would construct the parallelogram with straightedge and compass

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AFB and EFC have the same area (click below)

Parallelogram and similar triangles question

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