Computing area of a region bound by a triangle and two of its cevians via Menelaus Theorem

Question

$AD$ 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.

Answer 1

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, $[CDPE] = 19/2$.