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$AD$ and $BE$ are two cevians of $\triangle{ABC}$, and $P$ is the intersection of them. If the area of $\triangle\mathit{ABP}$ is 6, the area of $\triangle\mathit{AEP}$ is 3, and the area of $\triangle\mathit{BDP}$ is 4, compute the area enclosed by quadrilateral $\mathit{CDPE}$. The application of Meneaus' Theorem is suggested.

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Let $[CDPE] = x$. We may deduce the following through area ratios.

  • $EP/PB = 3/6 = 1/2$
  • $BD/DC = (6+4)/(3+x) = 10/(3+x)$
  • $CA/AE = (4+6+3+x)/(6+3) = (13+x)/9$

Truthfully, we don't even need Menelaus (Look! The purpose of Menelaus is to abstract away from the triangle).

Anyways, by Menelaus we have $$\frac{1}{2} \cdot \frac{10}{3+x} \cdot \frac{13+x}{9} = 1$$ Solving, $[CDPE] = 19/2$.

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    I would like to clarify your computation of the ratio $\left\vert EP \right\vert : \left\vert BP \right\vert$. –  Apr 26 '18 at 14:00
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    If the distance between $A$ and $\overline{BE}$ is $h$, $\frac{\left\vert \overline{\mathit{EP}} \right\vert}{\left\vert \overline{\mathit{BP}} \right\vert} = \frac{\frac{1}{2} \left\vert \overline{\mathit{EP}} \right\vert h}{\frac{1}{2} \left\vert \overline{\mathit{BP}} \right\vert h} = \frac{\left\vert \triangle\mathit{AEP} \right\vert}{\left\vert \triangle\mathit{ABP} \right\vert} = \frac{3}{6} = \frac{1}{2}$. –  Apr 26 '18 at 15:21
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    Is that correct? –  Apr 26 '18 at 15:23
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    You are applying Menelaus' Theorem to $\triangle\mathit{BCE}$ and transversal $\overline{\mathit{AD}}$. –  Apr 26 '18 at 16:36
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    @Adelyn Yes, that is correct. –  Apr 26 '18 at 18:46