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.  

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 $v = \dfrac {xyz +z^2x}{y^2 - xz}$
∴ $[\triangle ABC] = x + y + z + u + v = x + y + z + \dfrac {xyz +x^2z}{y^2 - xz} + \dfrac {xyz +z^2x}{y^2 - xz} = ... = 1365$.
