jmc

Since the ratio $AN:AC$ equals the ratio $AP:AB$ (each is $1:2$) and $\angle A$ is common in $△APN$ and $△ABC$, then $△APN$ is similar to $△ABC$.

Since the ratio of side lengths between these two triangles is $1:2$, then the ratio of areas is $1:2^2=1:4$.

Thus, the area of $△ABC$ is $4\times 2=\boxed{8}\text{mbox}{cm}^2$.