Find the area of the triangle There are two points $N$ and $M$ on the sides $AB$ and $BC$ of the triangle $ABC$ respectively. The lines $AM$ and $CN$ intersect at point $P$. Find the area of the triangle $ABC$, if areas of triangles $ANP, CMP, CPA$ are $6,8,7$ respectively.  


From the hypothesis, it follows that


*

*$CN:CP = area(\triangle CAN):area(\triangle CPA)=13:7$

*$MP:MA =area(\triangle CMP):area(\triangle CMA)=8:15$


Applying Menelaus' Theorem to $\triangle APN$ with $B, M,C$ we have that
$$\frac{BN}{BA}\cdot \frac{CN}{CP}\cdot \frac{MP}{MA}=1,\tag{3}$$
(note that we use non-directed length here). So
$$\frac{area(\triangle ABC)}{area(\triangle BCN)}=\frac{BA}{BN}=\frac{CP}{CN}\cdot \frac{MA}{MP}=\frac{7}{13}\cdot\frac{15}{8}=\frac{105}{104}.$$
It follows that $$area(\triangle ABC)=105\times area(\triangle ACN)=\color{red}{1365}.$$
