Brazilian Contest - Geometry

Its not a solution. But I put protest against the very low point of this question. I think a 7% solver's question does not worth 10 points.

Let $R$ be the midpoint of $\overline{BC}$ . $\triangle NAG \sim \triangle CAR$ by SAS, so $\overleftrightarrow{NG}\parallel \overleftrightarrow{CR}$ , so $\overleftrightarrow{NM}\parallel \overleftrightarrow{CB}$ . Therefore, $\triangle NAM \sim \triangle CAB$ . Therefore $\triangle CAB$ is a right triangle with a right angle at $B$ .

The length of a vertex is $\frac{2}{3}$ of the distance from that vertex to the midpoint on the opposite side. In a right triangle, this is half of the hypotenuse. We are given that the hypotenuse is 12cm, so half of that is 6cm. Therefore, the distance that we are looking for is $\boxed{4\text{cm}}$ .