What is the area of BEC shaded triangle?

Question

Area of $ABC$ triangle is $24\,\mathrm{cm}^2,$ $\frac{BM}{AB}=\frac{1}{3}$ and $\frac{AN}{AC}=\frac{1}{3}$. E is a midpoint of MN. What is the area of $BEC$ shaded triangle?

I don't know how to calculate the area of the shaded triangle, All I can think of is using triangle similarity or some sort of ratios with area/sides but I can't find it

Answer 1

Drop perpendiculars from $M$, $E$ and $N$ to $BC$ lets call their intersections by $X$, $Y$ and $Z$ respectively. $MNZX$ is a trapezoid and $EY$ is its' midline so: $$EY=\frac{MX+NZ}{2}$$ So if by $[F]$ we denote the area of a shape $F$ we get: $$[BEC]=\frac{[BMC]+[BNC]}{2}=\frac{\frac{1}{3}[ABC]+\frac{2}{3}[ABC]}{2}=\frac{1}{2}[ABC]=12$$ Where in the middle we made use of the fact that for the triangles that share the altitude the ratio of their areas is the same as ratio of their sides.

Answer 2

If $[\triangle BEM]=a$, then $[\triangle AEN]=[\triangle AEM]=2a$.

$[\triangle CEN]=4a$ and $[\triangle ABC]=4a\times\dfrac32\times3=18a$.

$18a=24$ $\implies $ $a=\dfrac43$.

$[\triangle BEC]=18a-9a=12$.