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.  

Let 
$$
\begin{cases}
a &= \verb/Area/(ANP) = 6\\
b &= \verb/Area/(CPA) = 7\\
c &= \verb/Area/(CMP) = 8
\end{cases}$$
and $(\alpha,\beta,\gamma)$ be the barycentric coordinate of $P$ with respect to $\triangle ABC$. i.e
the triplet of numbers such that
$$\vec{P} = \alpha \vec{A} + \beta \vec{B} + \gamma \vec{C}\quad\text{ subject to }\quad \alpha + \beta + \gamma = 1$$
Following is a picture illustrating the arrangement of points and labeling
(not drawn to scale because the actual triangle is highly acute and
hard to display)
$\hspace1in$ 
Since
$$
\begin{align}
\alpha &= \frac{\verb/Area/(CMP)}{\verb/Area/(CMA)} = \frac{c}{c+b}\\
\gamma &= \frac{\verb/Area/(ANP)}{\verb/Area/(ANC)} = \frac{a}{a+b}\\
\end{align}
\quad\text{ and }\quad
\frac{\verb/Area/(CPA)}{\verb/Area/(ABC)} = \beta = 1 - \alpha - \gamma
$$
This leads to $$\verb/Area/(ABC) = \frac{\verb/Area/(CPA)}{1 - \frac{c}{c+b} - \frac{a}{a+b}}
= \frac{b}{1 -\frac{c}{c+b} - \frac{a}{a+b}}
= \frac{b(a+b)(c+b)}{b^2-ac}\\ = \frac{7(6+7)(7+8)}{7^2-6\cdot 8} = 1365$$
