30. [14] Alice has an equilateral triangle $A B C$ of area 1. Put $D$ on $B C, E$ on $C A$, and $F$ on $A B$, with $B D=D C, C E=2 E A$, and $2 A F=F B$. Note that $A D, B E$, and $C F$ pass through a single point $M$. What is the area of triangle $E M C$ ?
Answer: $\frac{1}{6}$ Triangles $A C F$ and $B C F$ share a height, so the ratio of their areas is $A F / B F=1 / 2$. By the same method, the ratio of the areas of $A M F$ and $B M F$ is $1 / 2$. So, the ratio of the areas of
$A C M$ and $B C M$ is also $1 / 2$. Similarly, the ratio of the areas of $A B M$ and $B C M$ is $1 / 2$. But the sum of the areas of $A C M, B C M$, and $A B M$ is 1 , so the area of $A C M$ is $\frac{1}{4}$. Then the area of $E M C$ is $2 / 3$ the area of $A C M$, because they share heights, so their areas are in the same ratio as their bases. The area of $E M C$ is then $\frac{2 \cdot 1}{3 \cdot 4}=\frac{1}{6}$.