It is a very easy problem once you know the technique called mass-point geometry (or barycentric coordinates).

momo1729 wrote: Let $ABC$ be a triangle with area $1$ and $P$ the middle of the side $[BC]$. $M$ and $N$ are two points of $[AB]-\left \{ A,B \right \}$ and $[AC]-\left \{ A,C \right \}$ respectively such that $AM=2MB$ and$CN=2AN$. The two lines $(AP)$ and $(MN)$ intersect in a point $D$. Find the area of the triangle $ADN$. Well, since ratios of areas, and ratios of segments along a line are preserved by an affine transformation, therefore, it suffices to find the area of $\triangle ADN$ in an equilateral triangle $\triangle ABC$. Here, $AM = 2 \cdot MB, CN = 2 \cdot AN$ implies that $AM = 2 AN$. Combined with $\angle BAC = \angle MAN = 60^{\circ}$, this implies that $\triangle AMN$ is right angled at $N$. Now, $AP$ is the median $\implies$ it is also the angle bisector of $\angle BAC$, and thus $AD$ is the bisector of $\angle MAN$. Therefore, ($[.]$ denotes area) $$\frac{[ADN]}{[AMN]} = \frac{DN}{MN} = \frac{AN}{AN+AM} = \frac 13$$ . Combining this with $$[AMN] = \frac{[AMN]}{[ABC]} = \frac{AM}{AB} \cdot \frac{AN}{AC} = \frac 23 \cdot \frac 13 = \frac 29$$ , we finally get $[ADN] = \frac{2}{27}$.