Find the ratio of the areas of the triangles: $MPA$ and $PBC$

geometry area euclidean-geometry plane-geometry

We can use Menelaus' theorem to find the length $AP$. For the transversal $PMN$ and $\triangle ABC$, $$\frac{PA}{PC}\cdot \frac{CN}{NB} \cdot \frac{BM}{MA}=1$$ This gives $PA=6$. So $PA=AC \Rightarrow [BPA]=[BAC].$

Thus $$\frac{[MPA]}{[BPC]}=\frac{[MPA]}{[BPA]}\cdot\frac{[BPA]}{[BPC]}=\frac{MA}{BA}\cdot\frac{1}{2}=\frac{1}{5}.$$

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