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

Question

For reference:

The circle inscribed with a triangle ABC is tangent to $AB$ in $M$ ​​and to $BC$ in $N$. The prolongations of $NM$ and $CA$ intersect at $P$. Calculate the ratio of the areas of the triangles $MPA$ and $PBC$, if $AB = 5, BC = 7$ and $AC = 6$. Answer:($\frac{1}{5}$)

My progress:

$S_X = S{MBN},$ $S = S_{ABC},$ $S_Y=S_{AMNC}.$

$MB=x \implies 2(5-x)+2(7-x)+2x = 6+7+7$

$ \therefore x = 3=BN$

$\therefore MA = 2~\text{and}~NC = 4$

$\frac{S}{S_{CPN}} = \frac{4.2}{CP.NP}=\frac{8}{NP.NC}$

$\frac{S_X}{S} = \frac{3.3}{5.7} = \frac{9}{35}\implies S_X=\frac{9S}{35}$

$S_Y = S - S_X = \frac{26S}{35}$

$\frac{S_{CPN}}{S_{BNP}}= \frac{4}{3}$

$\frac{S_X}{S_{AMP}} = \frac{3MN}{2MP}$

$\frac{S}{S_{ABP}}=\frac{6}{AP}$

$\frac{S_{AMP}}{S_{ACN}}=\frac{AP.MP}{CP.CN}$

I found several relationships but I was not able to equate...???

Answer 1

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