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For math team at school, we recently went over Ceva Formula since it will appear on our future meets. This problem is related to it(at least I think so).

In the picture below, D is on BC, E is on CA, and F is on AB and the three Cevians meet at P with the following ratios of division: $\frac{BD}{DC} = \frac{1}{2}$, $\frac{CE}{AE} = \frac{3}{2}$, $\frac{AF}{FB} = \frac{4}{3}$. If the area of the triangle ABC is 45, find the area of quadrilateral CEPD.

Triangle from the problem

My initial plan for this problem was to find the length of each segment inside of the triangle using mass points. I cannot do this because I do not know the lengths of the sides, just the ratios. Later, assuming the first step succeeded I wanted to used Heron's formula to calculate the areas of the two triangles in the quadrilateral and add them together to get the answer. Is it possible to do with what I described above? Can it be done with using Area ratios?

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Area ratios are just fine: $$\frac{[PAB]}{[PAC]}=\frac{DB}{DC}=\frac{1}{2},\quad\frac{[PAC]}{[PBC]}=\frac{FA}{FB}=\frac{4}{3}$$ lead to: $$ [PAB]:[PBC]:[PCA] = 2: 3 :4 $$ hence $$ [PAB]=\frac{2}{9}[ABC],\quad [PBC]=\frac{3}{9}[ABC],\quad [PAC]=\frac{4}{9}[ABC]$$ and $$ [PDC]=\frac{DC}{BC}[PBC]=\frac{2}{3}[PBC],\quad [PCE]=\frac{CE}{CA}[PAC]=\frac{3}{5}[PAC]$$ so: $$[CEPD]=[PDC]+[PCE] = \left(\frac{2}{3}\cdot\frac{1}{3}+\frac{3}{5}\cdot\frac{4}{9}\right)[ABC] = \color{red}{\frac{22}{45}}[ABC].$$


An interesting alternative approach is given by the following observation: affine maps preserve the ratio $\frac{[CEPD]}{[ABC]}$, hence we may assume without loss of generality $B=(-6,0)$, $D=(0,0)$, $C=(12,0)$, $A=(0,5)$, $E=(4,3)$, $P=\left(0,\frac{5}{3}\right)$ then compute $[ABC]$ and $[CEPD]$ through the shoelace formula. $\frac{AP}{PD}=2$ by Van Obel's theorem.

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