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.  
 A: 
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 $$area(\triangle ABC)=105\times area(\triangle ACN)=\color{red}{1365}.$$
A: 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 $$\verb/Area/(ABC) = \frac{\verb/Area/(CPA)}{1 - \frac{c}{c+b} - \frac{a}{a+b}}
= \frac{b}{1 -\frac{c}{c+b} - \frac{a}{a+b}}
= \frac{b(a+b)(c+b)}{b^2-ac}\\ = \frac{7(6+7)(7+8)}{7^2-6\cdot 8} = 1365$$
A: Simple Geometry Approach
Note that, with the same altitude, ratio of areas of two triangles is equal to the ratio of their bases.

Therefore, $\dfrac {x}{y} = \dfrac {NP}{PC} = \dfrac {u}{v + z}$ ….. (*)
Similarly, $\dfrac {z}{y} = \dfrac {v}{u + x}$ ….. (#)
After eliminating v from (*) and (#) and making u as subject, we have
$ u = \dfrac {xyz +x^2z}{y^2 - xz}$
Similarly, or by symmetry, we have $v = \dfrac {xyz +z^2x}{y^2 - xz}$
∴ $[\triangle ABC] = x + y + z + u + v = x + y + z + \dfrac {xyz +x^2z}{y^2 - xz} + \dfrac {xyz +z^2x}{y^2 - xz} = ... = 1365$.
A: While the other approaches (using Menelaos' theorem or barycentric coordinates) are very elegant, here is one which might work if you can't think of all the clever geometric theorems needed for those.
Coordinate-based approach
Since all you talk about is areas, the whole setup is invariant under area-preserving affine transformations. Therefore you may without loss of generality assume that your triangle has a right angle at $B$, and choose your coordinate system such that you have
$$B=(0,0)\quad C=(c,0)\quad A=(0,a)\quad M=(m,0)\quad N=(0,n)$$

Here $c,a,m,n$ are the distance of the corresponding upper-case points from $B$. So you have four unknowns and three equations, which in general doesn't lead to a unique solution. But that's because we still haven't fully determined the shape of the triangle, since we could vary it's aspect ratio while preserving its areas. So again without loss of generality, let's assume that $P=(p,p)$ lies on the line $x=y$. Now let's formulate the equations. The lines $AM$ and $CN$ have equations
$$
AM: ax + my = am \qquad CN: nx + cy = cn
$$
so if $P$ is to lie on both of them, you have
\begin{align*}
ap + mp &= am \\
np + cp &= cn \\
\tfrac12\cdot p\cdot(a-n) &= 6 = \operatorname{area}(ANP) \\
\tfrac12\cdot(c-m)\cdot p &= 8 = \operatorname{area}(CMP) \\
\tfrac12\cdot(ac-ap-pc) &= 7 = \operatorname{area}(CPA)
\end{align*}
That system of equations has two real solutions, which only differ by a global change of all signs (i.e. by a rotation around the origin by $180°$). One of them is
\begin{align*}
a &= 15\sqrt{\frac{21}{2}} & a \approx 48.60555523805895 \\
c &= 26\sqrt{\frac{14}{3}} & c \approx 56.16641938620146 \\
m &= 60\sqrt{\frac{6}{7}} & m \approx 55.54920598635309 \\
n &= 52\sqrt{\frac{6}{7}} & n \approx 48.14264518817268 \\
p &= 4\sqrt{42} & p \approx 25.92296279363144
\end{align*}
$$ \operatorname{area}(ABC) = \tfrac12ac = 1365 $$
So in reality, your triangle looks more like this:

The role of orientation
But perhaps the question was meant in a different way. Perhaps $M$ and $N$ need not lie on the line segment between the corresponding triangle corners, but may lie anywhere on the infinite extension of that edge. Perhaps there is a reason why $ANP$ is oriented counter-clockwise in my first diagram but $CMP$ and $CPA$ are oriented clockwise. If I switch the sign of $CMP$ and $CPA$ (since counter-clockwise is usually considered the positive orientation) I get a different pair of solutions, which look like this:

Here $ANP, CMP, CPA$ are all oriented counter-clockwise. Here $\operatorname{area}(ABC)=-\frac{105}{97}$ which is negative since the triangle is oriented clockwise.
A: We may assume $P=(0,0)$, $A=(0,a)$, $C=(c,0)$, $N=(-v,0)$, and $M=(0,-u)$. The given information amounts to
$$ac=14,\qquad u={16\over c},\qquad v={12\over a}\ ;\tag{1}$$
furthermore the $x$-coordinate of $B=(C\vee M)\wedge(A\vee N)$ computes to
$$x_B=-{vc(a+u)\over ac-uv}\ .$$
The area of the triangle $\triangle:=\triangle(A,B,C)$ is then given by
$$|\triangle|={1\over2}(a+u)(c-x)={1\over2}{(a+u)(c+v)ac\over ac-uv}\ .$$
Using $(1)$ we get
$$|\triangle|={1\over2}\>{(ac+16)(ac+12)\over ac-{192\over ac}}=1365\ .$$
